# Prove or disprove $2abc(a+b+c)\ge 3(a^2b^2c^2+1)$

Let $a,b,c>0,ab+bc+ca=3$, prove or disprove $$2abc(a+b+c)\ge 3(a^2b^2c^2+1)$$

Now I can't find any counterexample

• What method are you using to prove it? A standard proof, by contradiction, also, It is great to show what you have done, if anything, so people don't think you are just saying 'solve this for me' – Chan Hunt May 26 '16 at 13:47
• How about $a = b = c = 1$ as a counterexample? – Aegon May 26 '16 at 14:03
• sorry,It's $a^2b^2c^2+1$ sorry – geromty May 26 '16 at 14:15
• Here's a counterexample for the corrected inequality: $(a,b,c) = (1,\frac52,\frac17)$. To find this, I fixed $a=1$ and plotted the region where the inequality holds, then looked for any point on the curve $b+c+bc=3$ outside that region. – Théophile May 26 '16 at 14:19
• For a counter eg, just allow $c\to 0$, the two remaining variables can satisfy the constraint and your inequality will read $0\ge3$. – Macavity May 26 '16 at 17:40

What follows is not so much an answer to the question (it was already fully answered in comments) as an essay intended to bring consolation to, perhaps even cheer up, the inequality stated in the question, which must feel somewhat mistreated, poor thing, and justifiably so.

The counterexample provided by "Macavity" is elegant $($albeit not fully spelled out$)$: decrease $c$ towards $0$ $($$c never reaches 0 since it must be strictly positive), fix a>0, and let b:=(3-ac)/(a+c) so that the constraint ab+ac+bc=3 is satisfied (also b>0 when c is small enough). Then abc\to a\cdot(3/a)\cdot 0 = 0, the LHS of the inequality converges to 0, the RHS converges to 3, thus for all small enough c the inequality LHS{}\geq{}RHS fails. The given inequality, which we were asked to prove or disprove, is definitely disproved. And that's it, problem solved. But is it? Have you gained any mental image of why the inequality fails to hold? Have you learned anything useful, anything that you may reuse, perhaps in a modified form, when working on a similar, or a not-so-similar problem? Have you at least acquired a couple of ways of looking for counterexamples, or of ways of making preliminar probes into a problem, trying to find its soft spots? My guess is that the anwers are, respectively, "no", "certainly not", and "very probably not". My experience as a budding mathematician (oh so long ago) was that the so called exercises are mere starting points, not ultimate goals. When doing an exercise I always kept asking additional questions which were as interesting, if not more interesting, than the exercise itself. Also, trying this or that approach I sometimes veered away from the theme of the exercise into an uncharted new territory; or during the work on the exercise there emerged an object with special properties, which made my analyze the object, perhaps study the class of all objects with those properties; or I took a second look at some failed attempt and noticed that the faulty reasoning I used in some situation may become a legitimate reasoning in a slightly different situation. And so on, and so forth. Now and then a work on an exercise developed into a miniature research project. And that's how I learned doing mathematics, working through some thousands of exercises in this peculiar way. Let us play around with the inequality in the present question and see where it leads us. The user "geromty", who asked the question, first gave the inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3(a^2b^2c^2+3) \end{equation*} (assuming a,b,c>0 and ab+ac+bc=3$$)$, which after the easy counterexample $a=b=c=1$ he changed to the present inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3(a^2b^2c^2+1)~. \end{equation*} Both versions of the inequality can be written as \begin{equation*} 2abc(a+b+c) \,\geq\, 3a^2b^2c^2 + \mathit{const}~, \end{equation*} the first one with $\mathit{const}=9$, the second one with $\mathit{const}=3$. Let us make another try with $\mathit{const}=0\,$: something tells us that third time round we might be lucky. Since $abc>0$, the inequality $2abc(a+b+c)\geq 3a^2b^2c^2$ is equivalent to the simpler inequality \begin{equation*} 2(a+b+c) \,\geq\, 3abc~. \end{equation*} Before we proceed, we define the function \begin{equation*} f_1(a,b,c) \,:=\, 2(a+b+c)-3abc~, \qquad a,b,c>0\,,~ ab+ac+bc = 3\,, \end{equation*} and ask Mathematica to minimize $f_1(a,b,c)$, within the given constraints: Wow! This is even better than what we expected: we have the inequality \begin{equation} 2(a+b+c) \,\geq\, 3(abc+1) \qquad \text{if $a,b,c>0$ and $ab+ac+bc = 3$}\,. \tag{1} \end{equation} If we multiply this inequality by $abc>0$, we obtain the $($valid$)$ inequality \begin{equation*} 2abc(a+b+c) \,\geq\, 3abc(abc+1)~, \end{equation*} with the same constraints on $a$, $b$, $c$ as inequality $(1)$, which suspiciously closely resembles the inequality in the question asked by "geromty": it is as if someone too hastily multiplied $abc+1$
by $abc$ and obtained $a^2b^2c^2+1$, never noticing the error.

Mathematica told us that the inequality $(1)$ is true, but that does not absolve us from actually proving it. We define the function \begin{equation*} f_2(a,b,c) \,:=\, 2(a+b+c) - 3(abc+1)\,, \qquad a,b,c>0\,,~ ab+ac+bc = 3~, \tag{2} \end{equation*} that is, $f_2(a,b,c)=f_1(a,b,c)-3$. Let us express, from the constraint $ab+ac+bc=3$, the variable $c$ as a function of the variables $a$ and $b$, \begin{equation*} c \,=\, \frac{3-ab}{a+b}~; \tag{3} \end{equation*} since $a+b>0$, the condition $c>0$ is equivalent to $ab<3$. We feed $c$ expressed by $a$ and $b$
as in $(3)$ to $f_2(a,b,c)\,$: \begin{equation*} g_2(a,b) \,:=\, f_2\Bigl(a,\,b,\,\frac{3-ab}{a+b}\Bigr)~, \qquad a,b>0\,,~ab<3\,. \end{equation*} We want to get a 'feel' for the behavior of the function $g_2$. The awkward thing about the funtion $g_2$ is that it is defined on an unbounded domain, namely on the region between the positive halves of the axes $(a)$ and $(b)$ and the branch of the rectangular hyperbola $ab=3$ in the first quadrant, and besides that it can attain arbitrarily large positive values. We want to actually see the function $g_2$
in its entirety, over all of its domain and over its complete range of values. This can be done.
The homeomorphism $\theta\colon\mathbb{R}_{\geq0}\to[\mspace{2mu}0,1)$ defined by \begin{equation*} \theta(x) \,:=\, \frac{x}{1+x}~, \qquad x \geq 0\,, \end{equation*} has the inverse \begin{equation*} \theta^{-1}(y) \,=\, \frac{y}{1-y}~, \qquad 0 \leq y < 1~. \end{equation*} We can extend the homeomorphism $\theta$ to a homeomorphism $[0,\infty]\to[0,1]$, still denoted $\theta$, by setting $\theta(\infty):=1$. To the points $0$, $1$, $\infty$ of the nonnegative half $[0,\infty]$ of the extended real axis there correspond, via the homeomorphism $\theta$, the points $0$, $1/2$ and $1$ of its "foreshortened"
image $[0,1]$. Now we can squeeze the infinite extents of the positive half-axes and of the range of $g_2$ into unit intervals and obtain a function \begin{equation*} h_2(u,v) \,:=\, \theta\bigl(g_2\bigl(\theta^{-1}(u),\,\theta^{-1}(v)\bigr)\bigr)~, \qquad\quad 0<u,v<1\,,\quad\frac{u}{1-u}\,\frac{v}{1-v}<3\,, \end{equation*} whose diagram within the confines of the unit cube represents, point by point, the entire diagram of the function $g_2$: Though the diagram is of the function $h_2$ of arguments $u$ and $v$, we read it as a diagram representing the function $g_2$ of arguments $a$ and $b$, with the tick's labels on the axes indicating the values of $g_2$ and its arguments. The function $h_2$ is defined on the set of all $(u,v)$ satisfying the inequalities $0<u<1$, $0<v<1$ and $v<(3-3u)/(3-2u)$; the part of the boundary of the domain of $h_2$ given by $v=(3-3u)/(3-2u)$ for $0\leq u\leq 1$ is an arc of the rectangular hyperbola $(u-3/2)(v-3/2)=3/4$. The two crossed lines drawn in the domain of $h_2$ intersect at the point $(u,v)=(1/2,1/2)$ that correponds to the one and only point $(1,1)$ at which the function $g_2$ attains its global minimum $0$.

It is high time we start thinking about actually proving the inequality \begin{equation*} g_2(a,b) \,\geq\, 0 \qquad \text{if $a,b>0$ and $ab<3$}\,, \tag{4} \end{equation*} which is equivalent to the inequality $(1)$ we intend to prove. We multiply the inequality $(4)$ by $a+b>0$, obtaining another inequality $G_2(a,b)>0$ equivalent to $(1)$, where $G_2(a,b):=(a\!+\!b)\mspace{-2mu}\cdot\mspace{-2mu}g_2(a,b)$. In the expanded form the inequality $G_2(a,b)>0$ reads \begin{equation*} \begin{split} 6 - 3a - 3b + 2a^2 - 7ab + 2b^2 + 3a^2b^2 &\,\geq\, 0 \\[.5ex] &\text{if $a,b>0$ and $ab<3$}\,. \end{split} \tag{5} \end{equation*} Let $\alpha$ be a real parameter in the range $0<\alpha<2$. We rewrite the left hand side of the inequality $(5)$ as \begin{align*} &6 \,+\, \alpha(a^2\mspace{-2mu}-\mspace{-2mu}2ab\mspace{-2mu}+\mspace{-2mu}b^2) \\[.5ex] &\qquad\quad {}\,+\, \bigl((2\!-\!\alpha)a^2\mspace{-2mu}-\mspace{-2mu}3a\bigr) \,+\, \bigl((2\!-\!\alpha)b^2\mspace{-2mu}-\mspace{-2mu}3b\bigr) \,+\, \bigl(3(ab)^2\mspace{-2mu}-\mspace{-2mu}(7\!-\!2\alpha)ab\bigr) \\[1ex] &\quad\,=\, 6 \,-\, \frac{9}{4(2\!-\!\alpha)} \,-\, \frac{9}{4(2\!-\!\alpha)} \,-\, \frac{(7\!-\!2\alpha)^2}{12} \,+\, \alpha(a\!-\!b)^2\\ &\qquad\quad \,+\, (2\!-\!\alpha)\Bigl(a\mspace{-1mu}-\mspace{-1mu}\frac{3}{2(2\!-\!\alpha)}\Bigr)^{\!2}\! \,+\, (2\!-\!\alpha)\Bigl(b\mspace{-1mu}-\mspace{-1mu}\frac{3}{2(2\!-\!\alpha)}\Bigr)^{\!2}\! \,+\, 3\Bigl(ab\mspace{-1mu}-\mspace{-1mu}\frac{7\!-\!2\alpha}{6}\Bigr)^{\!2}~. \end{align*} The constant part of the second formula $($it being constant means it does not depend on $a$ or $b$$), which is \begin{equation*} C(\alpha) \,:=\, 6 \,-\, 2\cdot\frac{9}{4(2\!-\!\alpha)} \,-\, \frac{(7\!-\!2\alpha)^2}{12}~, \end{equation*} cannot be >0 for any \alpha in the range 0<\alpha<2 since this would imply that there is an \varepsilon>0 such that g_2(a,b)\geq\varepsilon over all of the domain of g_2, which we know is not true because the global minimum of g_2 is 0 (we trust Mathematica in this). The best we may hope for is that C(\alpha)=0 for some \alpha within its prescribed range. Well, C(\alpha) is a quotient of polynomials, concretely it is a cubic polynomial in \alpha (with rational coefficients) divided by 2-\alpha. If the cubic polinomial in the numerator has a zero, it must have a double zero, so it will factor into linear factors. So let's see if our hunch is worth its salt: \begin{equation*} C(\alpha) \,=\, \frac{(\alpha-8)(2\alpha-1)^2}{12(2-\alpha)}~. \end{equation*} We are through, since we have C(\alpha)=0 by choosing \alpha=1/2, and with this value of \alpha the left hand side G_2(a,b) of (5) gets rewritten as \begin{equation*} G_2(a,b) \,=\, \tfrac{1}{2}(a\mspace{-2mu}-\mspace{-2mu}b)^2 +\, \tfrac{3}{2}(a\mspace{-2mu}-\mspace{-2mu}1)^2 +\, \tfrac{3}{2}(b\mspace{-2mu}-\mspace{-2mu}1)^2 +\, 3(ab\mspace{-2mu}-\mspace{-2mu}1)^2~. \end{equation*} Observe that G_2(a,b)\geq 0 for all real a and b, not just for a,b>0 satisfying ab<3, and that G_2(a,b)=0 if and only if (a,b)=(1,1). The fact that G_2(a,b)>0 for all a,b\in\mathbb{R} does not mean that g_2(a,b)>0 for all a,b\in\mathbb{R}. First of all, if a+b=0, then g_2(a,b) is not defined. And if a+b<0, then g_2(a,b)=G_2(a,b)/(a\!+\!b)<0 because G_2(a,b)>0. Now we are going to turn the tables: starting with the inequality (1), which we have just proved, we will concoct one truly remarkable inequality. You will have the rare opportunity to observe at close quarters the working of a devious mind of the kind that produces all those bothersome inequalities that you are then given to prove (or disprove). Let us introduce the set H_3:=\{(a,b,c)\in\mathbb{R}^3\mid a b+a c+b c=3\} (we should have done this the moment we first mentioned the constraint ab+ac+bc=3\,; but, as it is said, better late than never). If (a,b,c)\in H_3, then a+b\neq 0, and by symmetry also a+c\neq 0 and b+c\neq 0\,; indeed, a+b=0 implies -a^2 = ab+ac+bc = 3, which cannot be. Thus we can have either a+b>0 or a+b<0\,. This result is slightly unsettling: the condition (a,b,c)\in H_3 is symmetric in a, b, c while the condition a+b>0 is slanted towards the pair a, b. However, contrary to the appearance the condition a+b>0 is actually symmetric in a, b, c since it implies \begin{equation*} a+c \,=\, a+(3\!-\!ab)/(a\!-\!b) \,=\, (3\!+\!a^2)/(a+b) \,>\, 0~; \end{equation*} by symmetry the three conditions a+b>0, a+c>0, b+c>0 imply each other (supposing (a,b,c) is a point in H_3, do not forget about this). Similarly the conditions a+b<0, a+c<0, b+c<0 imply each other. Let us visualize the set H_3, over the whole of its extent. As with visualizing the function g_2(a,b) (for a,b>0 and ab<3$$)$, we have to squeeze an infinite extent into a finite interval, but now we have in each of the three cooordinate directions a two-way infinite extent, that of the whole real line, to squeeze. We achieve this by extending our 'squeeze' map $\theta\colon[0,\infty]\to[0,1]$ to a map $[-\infty,\infty]\to[-1,1]$, still denoted $\theta$, by setting $\theta(x):=-\theta(-x)$ for $-\infty\leq x\leq 0$. That is, for
a real $x\leq 0$ we have $\theta(x)=x/(1-x)$, and $\theta(-\infty)=-1$. The inverse of the extended $\theta$ gives $\theta^{-1}(y)=y/(1+y)$ for $-1<y\leq 0$ and $\theta^{-1}(-1)=-\infty$. The extended 'squeeze' map $\theta$ is constructed as a non-smooth spline of two smooth halves. This would be a serious shortcoming for theoretical 'squeezing' of, say, smooth fields or smooth differential forms, but it is good enough for our visualizations. If you insist on using an overall smooth homeomorphism $(-\infty,\infty)\to(-1,1)$, and one that is an odd functions to boot, then there is no shortage of such maps. Here are two maps for you to choose from: one is $x\mapsto (2/\pi)\arctan(x)$ with the inverse $y\mapsto\tan(\pi y/2)$, another one is $x\mapsto (4/3)x/\bigl(1\!+\!\sqrt{1\!+\!((4/3)x)^2}\,\bigr)$ with the inverse $y\mapsto (3/2)y/\bigl(1\!-\!y^2\bigr)\,$. The curious multipliers $4/3$ and $3/2$ appearing in the formulas for the second map are there to ensure that
the points $-1$ and $1$ of the real line correspond to the points $-1/2$ and $1/2$ in the interval $(-1,1)$.

From the defining equation $ab+ac+bc=3$ of the set $H_3$ we express $c\in\mathbb{R}$ as a function $c(a,b)=(3\!-\!ab)/(a\!+\!b)$ of $a,b\in\mathbb{R}$, $a+b\neq 0$. In order to see the set $H_3$ using our infini-vision, we draw the diagram of $w(u,v):=\theta\bigl(c\bigl(\theta^{-1}(u),\theta^{-1}(v)\bigr)\bigr)\in(-1,1)$, where $u,v\in(-1,1)$, $u+v\neq 0$.$\,$ $($Because $\theta$ is an odd function, the points $a$ and $b$ of the real line such that $a+b=0$ correspond to the points $u:=\theta(a)$ and $v:=\theta(b)$ in the interval $(-1,1)$ such that $u+v=0$.$)\,$ Here it is, the diagram: Note that the points on the diagonals forming the boundary of the diagram, such as the diagonal from $(-\infty,\infty,\infty)$ to $(\infty,-\infty,\infty)$, do not belong to the diagram proper. We clearly see that the parallel projection of the set $H_3$ to the coordinate plane $(a,b)$ along the axis (c) is the whole plane $\mathbb{R}^2$ minus the 'cross-diagonal' $\{(x,-x)\mid x\in\mathbb{R}\}$, and that the same is true of the projection to the plane $(a,c)$ along the axis (b) and of the projection to the plane $(b,c)$ along the axis (a). And here is the part of the set $H_3$ in the positive octant $\mathbb{R}_{>0}^3$, with which we began the discussion of the inequality $(1)$: Now we extend the function $f_2$, which was defined in $(2)$ only on the part of $H_3$ in the positive octant, to the whole $H_3$: \begin{equation*} f_2(a,b,c) \,:=\, 2(a+b+c) - 3(abc+1)~, \qquad ab+ac+bc = 3~. \tag{6} \end{equation*} Let us also recall the definitions of the elementary symmetric polynomials \begin{align*} s_1 \,=\, s_1(a,b,c) &\,:=\, a+b+c~, \\[.5ex] s_2 \,=\, s_2(a,b,c) &\,:=\, ab+ac+bc~, \\[.5ex] s_3 \,=\, s_3(a,b,c) &\,:=\, abc~, \end{align*} which we at once employ to recast the definition $(6)$ of the extended function $f_2$, \begin{equation*} f_2(a,b,c) \:=\, 2s_1(a,b,c) - 3s_3(a,b,c) - 3~, \qquad s_2(a,b,c) = 3~. \tag{7} \end{equation*} We introduce the function $\overline{f}_2(a,b,c):=-f_2(-a,-b,-c)$ for $s(a,b,c)=3$ $($note that $s(a,b,c)=3$ implies s(-a,-b,-c)=s(a,b,c)=3), so that \begin{equation*} \overline{f}_2(a,b,c) \,=\, 2s_1(a,b,c) - 3s_3(a,b,c) + 3~, \qquad s_2(a,b,c)=3~. \tag{8} \end{equation*} We have \overline{f}(a,b,c)\geq 6 when s_2(a,b,c)=3 and a+b>0. The function \begin{equation*} f_3(a,b,c) \,:=\, f_2(a,b,c)\,\overline{f}_2(a,b,c)~, \qquad s_2(a,b,c) = 3~, \tag{9} \end{equation*} which is defined on H_3, satisfies the condition that f_3(-a,-b,-c)=f_3(a,b,c) for every (a,b,c)\in H_3, and it follows that \begin{equation*} f_3(a,b,c)\geq 0 \qquad \text{everywhere on H_3}\,, \end{equation*} where f_3(a,b,c)=0 if and only if (a,b,c)=(1,1,1) or (a,b,c)=(-1,-1,-1). Using (7) and (8) we rewrite (9) as \begin{equation*} f_3 \,=\, (2s_1-3s_3)^2 - 9~, \qquad \text{defined when s_2=3}\,. \tag{10} \end{equation*} We introduce yet another function, \begin{equation*} f_4 \,:=\, (2s_1s_2-9s_3)^2 - 3s_2^3~, \tag{11} \end{equation*} which is a homogeneous polynomial in a, b, c of degree 6 defined everywhere on \mathbb{R}^3. We claim that \begin{equation*} f_4(a,b,c) \,\geq\, 0 \qquad \text{for all a,b,c\in\mathbb{R}}~. \tag{12} \end{equation*} This is the truly remarkable inequality we have promised a while ago. Fully expanded the inequality (12) reads \begin{equation*} \begin{aligned} &4\bigl(a^4 b^2 + a^2 b^4 + a^4 c^2 + a^2 c^4 + b^4 c^2 + b^2 c^4\bigr) \\ &\qquad {} \,+\, 8\bigl(a^4 b c + a b^4 c + a b c^4\bigr) \,+\, 5\bigl(a^3 b^3 + a^3 c^3 + b^3 c^3\bigr) \,+\, 15 a^2 b^2 c^2 \\ &\qquad\qquad {} \,-\, 13\bigl(a^3 b^2 c + a^3 b c^2 + a^2 b c^3 + a^2 b^3 c + a b^3 c^2 + a b^2 c^3\bigr) ~\geq~ 0~. \end{aligned} \end{equation*} Proving the inequality (12) is simple. If s_2(a,b,c)\leq0, then it is evident from (11) that f_4(a,b,c)\geq 0. Now the case s_2(a,b,c)>0. We first observe that s_2=3 implies that f_3 is defined, and that f_4=9f_3 which is clear from (10) and (11). Then with \kappa:=\sqrt{3/s_2(a,b,c)} we have s_2(\kappa a,\kappa b,\kappa c) = \kappa^2 s_2(a,b,c) = 3, and \begin{equation*} \kappa^6 f_4(a,b,c) \,=\, f_4(\kappa a,\kappa b,\kappa c) \,=\, 9 f_3(\kappa a,\kappa b,\kappa c) \,\geq\, 0~, \end{equation*} therefore f_4(a,b,c)\geq 0.~ Done. The zeros of f_4 are easy to determine. If s_2(a,b,c)>0, then f_4(a,b,c)=0 iff f_3(\kappa a,\kappa b,\kappa c) = 0 (with \kappa as above), iff (\kappa a,\kappa b,\kappa c)=\pm(1,1,1), iff a=b=c (where the common value of a, b, c is \neq 0 since s_2(0,0,0)=0). Otherwise, if $s_2(a,b,c)\leq 0$, then $f_4(a,b,c)=0$ iff $s_2(a,b,c)=0$ and $-9s_3(a,b,c)=2s_1(a,b,c)s_2(a,b,c)-9s_3(a,b,c)=0$, iff at least two of $a$, $b$, $c$ are $0$. Here's why: because of $abc = s_3(a,b,c)=0$ at least one of $a$, $b$, $c$ is $0\,$; if, say, $a=0$, then $bc=s_2(0,b,c)=0$, hence $b=0$ or $c=0$. Conversely, if at least two of $a$, $b$, $c$ are $0$, then $s_2(a,b,c)=0$ and $s_3(a,b,c)=0$, whence $f_4(a,b,c)=0$. The set of zeros of $f_4$ is therefore the union of the one-dimensional subspaces $\mathbb{R}u$ of $\mathbb{R}^3$ for $u\in \{(1,1,1),\,(1,0,0),\,(0,1,0),\,(0,0,1)\}$. In other words, the homogeneous polynomial $f_4$ has besides the trivial 'affine' zero $(0,0,0)$
$($which is a zero of every nonconstant homogeneous polynomial$)$ the set $Z_4=\{(1\!:\!1\!:\!1),\,(1\!:\!0\!:\!0),\,(0\!:\!1\!:\!0),\,(0\!:\!0\!:\!1)\}$ of four 'projective' zeros.

Now we of course want to see what the function $f_4$ looks like. It helps that $f_4(a,b,c)$ is a homogeneous polynomial in $(a,b,c)$; if we know its values on the unit sphere $S^2 = \{(a,b,c)\mid a^2+b^2+c^2=1\}$ $($the "$2$" in $S^2$ is a superscript, not an exponent$)$, then we know its values on a sphere $rS^2$ of any radius $r\,$: every point in $rS^2$ is of the form $ru$ for some $u\in S^2$
and therefore $f_4(ru) = r^6 f_4(u)$ (where the exponent $6$ is the degree of $f_4$). An attractive way to represent values of some nonnegative function $h$ defined on the unit sphere is the one used for the visualization of the Earth's geoid: at each point $u\in S^2$ we represent the value $h(u)$ by the point $(1+\rho\mspace{1mu}h(u))\mspace{1mu}u$, that is, by the point at the elevation $\rho\mspace{1mu}h(u)$ vertically above the point $u$ on the sphere, where "vertical" means "in the direction of the vector $u$" $($which points straight up$)$.
The positive constant $\rho$ is a suitably chosen scaling factor: usually we will want the average value of $\rho\mspace{1mu}h$ on the unit sphere to be a fraction of its radius $1$, that is, some value between $0$ and $1$, not too small and not too large, and we choose a factor $\rho$ accordingly. Of course, if you like the 'geoid' extending halfway to the Moon, or, at the other extreme, to be barely discernible from the unit sphere, with microscopic maximum heights, go forth and choose a scaling factor to achieve the appearance your heart desires.

Let us ask Mathematica to tell us the maximum value of $f_4$ on the unit sphere: With the maximum slightly over $2$ the scaling factor $\rho=1/3$ seems reasonable. Here is the corresponding geoid of $f_4$: We oriented the geoid so that the vector $(1,1,1)$ points upwards. The blue dot at the top marks the position of the zero $(1,1,1)/\sqrt{3}$ of $f_4$ on the unit sphere. The polynomial $f_4$ has eight zeros on the unit sphere $($each of the four projective zeros of $f_4$ is represented on the unit sphere by a pair of antipodal affine zeros$)$, but only three of them are visible.

Here's a thought: why not make an anti-geoid of a function $h$ defined on the unit sphere, representing the value of $h$ at a point $u\in S^2$ by the point $(1-\rho\mspace{1mu}h(u))\mspace{1mu}u$ below the point $u$ on the sphere. If we choose the scaling factor $\rho$ so that the maximum value of $\rho\mspace{1mu}h(u)$ will be less than $1$, then we will obtain a non-self-intersecting surface in the interior of the unit sphere. Let's do this
for $f_4$. We choose $\rho=1/3$ as for the geoid; but this time we draw only the upper hemisphere and the part of the anti-geoid of $f_4$ it contains, and make everything, the hemisphere and the anti-geoid inside it, transparent: As before, the blue dots mark the positions of zeros of $f_4$ on the unit sphere. The result is slightly reminiscent of snow globes $($well, of half-globes, without the snow$)$.

We shall use the inequality $(12)$ to produce another remarkable inequality \begin{equation*} f_5(a,b,c) \,\geq\, 0 \qquad \text{for all $a,b,c\in\mathbb{R}$}~, \tag{13} \end{equation*} where $f_5$ will be a symmetric homogeneous polynomial of degree $6$. To this end we first restrict the inequality $(12)$ to the strictly positive $a$, $b$, $c\,$: \begin{equation*} f_4(a,b,c) \,\geq\, 0 \qquad \text{for all $a,b,c>0$}~. \tag{14} \end{equation*} Into this restricted inequality we introduce new strictly positive variables $x:=bc$, $y:=ac$, $z:=ab$. $($There was an answer which employed this change of variables, but it was later removed by its author. With my reputation far below 10,000 I am not able to peek into the deleted answer in order to find out and tell you who posted it.$)$ The transformation $(a,b,c)\mapsto(x,y,z)$ is a bijection $\mathbb{R}_{>0}^3\to\mathbb{R}_{>0}^3$; the inverse transformation is $(x,y,z)\mapsto\bigl(\sqrt{yz/x},\,\sqrt{xz/y},\,\sqrt{xy/z}\,\bigr)$. From the inequality $(14)$ we therefore obtain the inequality \begin{equation*} f_4\Bigl(\sqrt{yz/x},\,\sqrt{xz/y},\,\sqrt{xy/z}\,\Bigr) \,>\, 0 \qquad \text{for all $x,y,z>0$}~. \tag{15} \end{equation*} The left hand side of this inequality is not a polynomial, it is a quotient $f_5(x,y,z)/(xyz)$, where $f_5(x,y,z)$ is a symmetric homogeneous polynomial of degree $6$. So we multiply the inequality $(15)$ by $xyz>0$, then rename the variables $x$, $y$, $z$ back to $a$, $b$, $c$, and thus obtain the promised inequality $(13)$, although restricted to $a,b,c>0$. It is a fact that $f_5$ satisfies the unrestricted inequality $(13)$, for any $a,b,c\in\mathbb{R}\,$: Mathematica assures us that that it is so (as we shall see shortly). At this point I have no idea how to go about proving the inequality $(13)$. Do you? Have an idea, I mean, how to prove $(13)$?

The symmetric polynomial $f_5$ of $a$, $b$, $c$ can be expressed as a polynomial in the elementary symmetric polynomials $s_1$, $s_2$, $s_3$, thus: \begin{equation*} f_5 \,=\, (2s_1s_2-9s_3)^2 - 3s_1^3s_3~. \tag{16} \end{equation*} The formula for $f_5$ differs very little from the formula $(11)$ for $f_4$: instead of the term $-3s_2^3$ in $(11)$ there is the term $-3s_1^3s_3$ in $(16)$.

Since the polynomial $f_5$ is symmetric and homogeneous of even degree, we only need to verify that $f_5(a,b,1)\geq 0$ for $-1\leq a,b\leq 1$ to be assured that $f_5(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$. Indeed, if $(a,b,c)\neq(0,0,0)$, one of the absolute values $|a|$, $|b|$, $|c|$ will be the largest and nonzero,
and because of the symmetry it suffices to consider the case where $|c|$ is the largest; but then $f_5(a,b,c)=c^6f_5(a/c,\,b/c,\,1)$, where $-1\leq a/c,\,b/c\leq 1$. So we make Mathematica minimize $f_5(a,b,1)$ on the square $-1\leq a,b\leq 1$: Therefore the inequality $(13)$ holds $($though we do not know how to prove it$)$. The point $(a,b)=(0,0)$ may not be the only point in the square $-1\leq a,b\leq 1$ at which $f_5(a,b,0)$ attains the minimum $0$, so we ask Mathematica to find all zeros of $f_5$ in the square: It follows that the set of all zeros of $f_5$ is the union of the seven one-dimensional subspaces $\mathbb{R}u$ for $u\in \{(1,1,1),\,(1,0,0),\,(0,1,0),\,(0,0,1),\,(1,-1,0)\,,(0,1,-1)\,,(-1,0,1)\}$. We also request the maximum value of $f_5$ on the unit sphere, which is only a little larger than the maximum value of $f_4$ on the unit sphere, so we again choose the scaling factor $\rho=1/3$ with which to construct the geoid of $f_5$, and also the anti-geoid of $f_5$ inside the unit hemisphere, As before, the blue dots mark the positions of zeros of $f_5$ on the unit sphere.

What do we get if we apply to the function $f_5$ the same change of variables that we applied to the function $f_4$ to obtain the function $f_5$? We are not very surprised when we find that we get back the function $f_4$. This is not a mere lucky coincidence:

Lemma 1. $~$Let $h(a,b,c)$ be a symmetric homogeneous polynomial of degree $6$ in real variables $a$, $b$, $c$, with the individual degree $\deg_a h$ at most $4$ $($and hence also $\deg_b h\leq 4$ and $\deg_c h\leq 4$$)$. Then the function $h^*(a,b,c)$ of real variables $a$, $b$, $c$, which is defined by \begin{equation*} h^*(a,b,c) \,:=\, abc\mspace{-1mu}\cdot\mspace{-2mu} h\bigl(\sqrt{b}\mspace{1mu}\sqrt{c}/\mspace{-1mu}\sqrt{a},\, \sqrt{a}\mspace{1mu}\sqrt{c}/\mspace{-1mu}\sqrt{b},\, \sqrt{a}\mspace{1mu}\sqrt{b}/\mspace{-1mu}\sqrt{c}\,\bigr)~, \end{equation*} is a symmetric homogeneous polynomial of degree $6$ in $a$, $b$, $c$ such that $\deg_a h^* \leq 4\,$. Moreover, $h^{**}=h\,$.

$($The polynomial $h^*(a,b,c)$ is computed in the field $\mathbb{R}\bigl(\sqrt{a},\sqrt{b},\sqrt{c}\,\bigr)$, where $a$, $b$, $c$ are formal indeterminates and $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$ are symbols whose squares are, respectively, $a$, $b$, $c$.$)$ The proof
of Lemma 1 is straightforward: just observe how the monomials are transformed by the change of variables -- there are only five monomials to consider.

The fact that $f_4^{**} = f_4$ leads to a simple proof of the inequality $(13)$. This is a special instance of
a more general result:

Lemma 2. $~$If $h$ and $h^*$ are as in Lemma 1, then $h(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$ if and only if $h^*(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}$.

Proof.\, Suppose that $h^*(a,b,c)\geq 0$ for all $a,b,c\in\mathbb{R}\,$. Applying the change of variables backwards we get $h(a,b,c)=h^*(bc,ac,ab)/(abc)^2$ for all $a,b,c\neq 0$, hence $h(a,b,c)\geq 0$ for all $a,b,c\neq 0\,$. Since the polynomial function $\mathbb{R}^3\to\mathbb{R} : (a,b,c)\mapsto h(a,b,c)$ is continuous, and the set of all $(a,b,c)\in\mathbb{R}^3$ such that $a,b,c\neq 0$ is a dense (open) subset of $\mathbb{R}^3$, it follows that $h(a,b,c)\geq 0$ for all $(a,b,c)\in\mathbb{R}^3$. The converse follows because $h^{**}=h$ by Lemma 1.$\,$ Done.

This is not the end of the story. The present essay is intended to be just the first one in a series of essays dedicated to the inequality stated in the question.

I'll be back.

Later. $~$Because of the unanimous protests (of one) against long and unwanted answers I decided that enough is enough and so won't be back, anytime soon, anywhere on StackExchange Mathematics. This is no web site for old mathematicians with a didactic leaning.

• Lol, please, don't be back, no more. But seriously, this answer is longer than appropriate for this site – Yuriy S Jun 3 '16 at 17:46
• @Yuriy - What's wrong with a long answer? True, it is not really an answer to the question as it was asked, which is barely alive. I hate seeing, times and again, hastily posted homework exercises, which are then hastily and mostly sloppily discussed in comments, and nobody, but nobody learns anything. This answer is long, but from it one can learn something, not only about the inequality stated in the question and its cousins, but about the environment in which such inequalities live. Is giving informative answers, albeit answers in an extended sense, against the spirit of SE mathematics? – chizhek Jun 3 '16 at 18:07
• I see nothing wrong with answers (or questions) of any length, but this site has certain rules, and it's not intended for article length posts – Yuriy S Jun 3 '16 at 18:09
• @Yuriy - Where are this rules? I looked through help on SE Mathematics and have seen no such rule. I asked on SE Mathematics Meta about upper limit on the size of answers and/or questions (they told me it is 30,000 characters) and nobody warned me that article length posts are no-no. Perhaps the rules are global, for the whole SE, not just the SE Mathematics. – chizhek Jun 3 '16 at 18:18
• Then no problem. But I've seen a lot of comments like 'this question requires an answer too long for Math.SE' – Yuriy S Jun 3 '16 at 18:21