# How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$

let $a,b,c>0$,and such $a+b+c=3$,

show that $$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$

I think this inequality use this $$ab\le\dfrac{(a+b)^2}{4}$$

• the best I could do was this: by AM $\ge$ HM, we have that: $\frac{3}{\dfrac{1}{(a+b)(4-ab)}+\dfrac{1}{(b+c)(4-bc)}+\dfrac{1}{(a+c)(4-ac)}} \leq \frac{(a+b)(4-ab)+(a+c)(4-ac)+(b+c)(4-bc)}{3}=(*)$ using $a+b+c=3$, $a+b=3-c$,$a+c=3-b$ and $c+b=3-a$ we have: $(*)=\frac{(3-c)(4-ab)+(3-b)(4-ac)+(3-a)(4-bc)}{3}=\frac{36-3(ab+ac+bc)-4(a+b+c)+3abc}{3}=(**)$ Using $a+b+c=3$: $(**)=\frac{24-3(ab+ac+bc)+3abc}{3} \leq \frac{24+3}{3}=9$ which gives $2/3$ instead of $1$ on the right side of the inequality, so, it's sort of close... maybe soneone can inprove that (or use HM-GM instead of HM-AM). – user119459 Jan 8 '14 at 15:14
• Since the function $f(x)=\frac{1}{(3-x)(4-K/x)}$, where positive, is convex, $$\sum_{cyc}\frac{1}{(3-c)(4-ab)}\geq\frac{3}{4-abc}\geq\frac{3}{4}$$ by the Jensen inequality. – Jack D'Aurizio Jan 8 '14 at 21:49
• @JackD'Aurizio: Have you tried your inequality for $a=b=c=1$? You obtain $1/2 \geq 3/4$... And your $K$ is not fixed, it depends on $a,b,c$, which doesn't make it a constant when trying to prove that your function is convex. – Beni Bogosel Jan 9 '14 at 13:02
• @Beni Bogosel: Sorry, I forgot a factor $2$ in the LHS. The correct statement is: $$\sum_{cyc}\frac{2}{(3-c)(4-ab)}\geq\frac{3}{4-abc}.$$ Using the Jensen inequality is not an overplay, since we can assume $a,b,c>0,a+b+c=3,abc=K\geq 1$ and prove the inequality with these constraints, exploting the convexity of $f(x)$. – Jack D'Aurizio Jan 9 '14 at 13:59
• @JackD'Aurizio: The function you consider is not convex on $(0,3)$ because it has a singularity at $x=k/4$ which is definitely in that interval since $K=abc \leq 1$. – Beni Bogosel Jan 9 '14 at 18:38

Here is a “brute-force” proof. Let us denote your sum by $S$. We may assume without loss that $a$ is smaller than both $b$ and $c$ ; then $a\leq 1$. Let us put $s=b+c=3-a,t=b-c$.

Lemma There are numbers $M,a_0,a_1,a_2,a_3,b_1,b_2,b_3$ that can be written as polynomials in $a$, such that $b_1,b_2,b_3$ are all nonnegative, $M$ is positive, $a_0=\frac{4M}{(6-s)(s^2-3s+8)},a_2=\frac{M}{2s(16-s^2)^2}$, and the identity $M-(4-ab)(a+b)(a_0+a_1t+a_2t^2+a_3t^3)=ct^2(b_1+b_2c+b_3c^2)$ holds.

The (ugly) proof of this lemma is deferred to the end of this answer. Noticing that when $b$ and $c$ are interchanged, $t$ becomes $-t$, and noticing also the identity $(16-s^2)^2-4(4-bc)(16-s^2-(b-c)^2)=(b-c)^4$ (remember that $s=b+c$), we see that

$$\begin{array}{lcl} \frac{1}{(a+b)(4-ab)} &\geq& \frac{a_0+a_1t+a_2t^2+a_3t^3}{M} \\ \frac{1}{(a+c)(4-ac)} &\geq& \frac{a_0-a_1t+a_2t^2-a_3t^3}{M} \\ \frac{1}{(b+c)(4-bc)} &\geq& \frac{4}{s(16-s^2)}-\frac{4}{s(16-s^2)^2}t^2 \ \\ \end{array}$$

Adding up and multiplying by $2$, we see that

$$S \geq \big(\frac{4a_0}{M}+\frac{8}{s(16-s^2)}\big)+ \big(\frac{4a_2}{M}-\frac{8}{s(16-s^2)^2}\big)t^2= \frac{8}{(6-s)(s^2-3s+8)}+\frac{8}{s(16-s^2)}$$

We are then done by noticing that $s\in[2,3]$ and that the RHS of this inequality can be rewritten as

$$1+\frac{4+(s-2)^2(2(3-s)+8)+(s-2)^3\big(\frac{3-s}{2}\big)}{s(6-s)(16-s^2)(s^2-3s+8)}$$

Equality is reached iff $a=b=c=1$.

Proof of lemma Take

$$\begin{array}{lcl} M &=& 3(((a-7)(a^2-9)(a+1))^2)(a^2-3a+4)((a^2-3a+8)^2) \\ a_1 &=& 24(((a-7)(a-3)(a+1))^2)(3a-4)(a^2-3a+4) \\ a_3 &=& (1-a)(a^8 - 14a^7 + 177a^6 - 1152a^5 + 3903a^4 - 1362a^3 - 10033a^2 + 5984a - 4416) \\ b_1 &=& (((1-a)^{10} + 26(1-a)^9 + 213(1-a)^8 + 1572(1-a)^7 + 7256(1-a)^6 + 18720(1-a)^5)a^2 + (19696(1-a)^5 + 91664(1-a)^4 + 58944(1-a)^3 + 23040(1-a)^2 + 6912(1-a))a + 6912(1-a)) \\ b_2 &=& 3a^{11} - 66a^{10} + 842a^9 - 7038a^8 + 36340a^7 - 100374a^6 + 89838a^5 + 134190a^4 - 316271a^3 + 249000a^2 - 107968a + 35328 \\ b_3 &=& a(1-a)( -2a^8 + 28a^7 - 354a^6 + 2304a^5 - 7806a^4 + 2724a^3 + 20066a^2 - 11968a + 8832) \\ \end{array}$$

• Yes, it's quite typical. With 90% of what one tries with answering this question, ugliness is just around the corner :-( – Han de Bruijn Jan 13 '14 at 16:06

Step 1. Transformation.
The three dimensional problem can be re-formulated as a two dimensional problem by introducing, quite like here , exactly the same triangle coordinates: $$\left[ \begin{array}{c} a \\ b \\ c \end{array} \right] = \left[ \begin{array}{c} 3 \\ 0 \\ 0 \end{array} \right] + \left[ \begin{array}{c} -3 \\ 3 \\ 0 \end{array} \right] x + \left[ \begin{array}{c} -3 \\ 0 \\ 3 \end{array} \right] y$$ Then the inequality $\;\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1\;$ does not so much "simplify", but anyway becomes an inequality in two variables. And the equation $\;a + b + c = 3\;$ corresponds with a normed 2-D triangle, with vertices $(0,0),(1,0),(0,1)$ .

Step 2. Visualization.
The insides and outsides of both can easily be visualized, as has been done in the above picture on the right (seen in the plane of the triangle, with a transformation rectangular $(x,y) \rightarrow$ equilateral) :
$\color{red}{red}$ for the edges of $\;a + b + c = 3\;$ and $\color{green}{green}$ for $\;\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)} < 1\;$ and simply white for $\;\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)} > 1\;$ . (With a smooth function - this one is not - the isolines would look a lot "neater".)
It is seen in the same picture that the vertices of the triangle could well have been put on the curve $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)} = 1$ . However, this would require that $x=0$ and/or $y=0$ , which is excluded by $\;b=3x,c=3y$ both $> 0$ . For the rest the triangle is seen to be on the outside of that curve , i.e. in the white area, thus establishing, at first sight, the given inequality.

Step 3. Triangle edges.
Due to the denominators which can be zero, singularities - $\color{blue}{blue}$ lines in the middle picture - are an issue with this problem. It is seen that the vertices of the triangle are located at singular spots. Without the additional requirement $a,b,c > 0$ that would have been problematic.
Indeed. Define the function: $$f(a,b,c) = \frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}$$ And specify for the triangle coordinates along one of the edges, say $c=3y=0$ , then: $$f(a=3-3x,b=3x,c=0) = \frac{2}{3[4-9(1-x)x]} + \frac{2}{12x} + \frac{2}{12(1-x)} \\ = \frac{2/27}{(x-1/2)^2+7/36} + \frac{1/6}{1/4-(x-1/2)^2}$$ As expected, singularities are found for $x=0,1$ . The minimum function value along a triangle edge equals $22/21 > 1$ and is found for $x=1/2$ . To be certain, by differentiation we find: $$f'(x) = -\frac{2(2x-1)}{27\left[(x-1/2)^2+7/36\right]^2} + \frac{2x-1}{6\left[1/4-(x-1/2)^2\right]^2} = 0$$ From which it is immediately clear that $x=1/2\;$ is a solution. And in the neighborhood $x\approx 1/2$ : $f'(x) \approx (-2/27(36/7)^2+4^2/6)(2x-1) = 104/147(2x-1)$ which is $< 0$ for $x < 1/2$ and $> 0$ for $x > 1/2$ , hence a minimum.

Step 4. Triangle inside.
There is not a whole bunch of points inside the triangle where $f(a,b,c) = 1$, because otherwise we would have seen them as green pixels. But the fact that they are not seen is not a proof that they do not exist. So we still have to establish, analytically, that $f(a,b,c)$ assumes no smaller values than $1$ inside the triangle.
Luck is on our path. In one of the comments accompanying the question Why does Group Theory not come in here? a key reference is mentioned:

From this reference we have the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$. On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.
This is precisely what we need. We have already defined the function $f(a,b,c)$ . Define the function $\;g(a,b,c) = a+b+c\;$ too. Then $\;g(r,r,r) = 3r = 3\;$ hence the point $P$ is: $\;(a,b,c) = (1,1,1)\;$ and $\;f(a,b,c) = 1\;$ (which is indeed a single "green" point, at the barycenter of the triangle). This value of $f$ cannot be a maximum because we have established greater $f$ - values along the triangle edges , e.g. $f(3/2,3/2,0) = 22/21$ . So it must be a minimum. This completes the proof.

Update. As a reply to the comments about whether there could perhaps exist a minimum at a lower level than $1$ , here comes a contour plot of $\;f(a,b,c)\;$ values in the plane of the triangle at levels $\; 1 < (2^k*10+2)/(2^k*10+1) \quad , \quad k=0 , \cdots , 11$ .
Showing a rather "flat" function behaviour, though definitely descending towards the middle of the triangle : the perimeters of the contours are invariably diminishing with increasing values of $\;k\;$.

Let's specify $f(a,b,c)$ for $y=x\;$ i.e. $\;a=3(1-2x) , b=3x , c=3x$ . This is the yellow line in the contour plot on the left. Then we have: $$f(x) := f(3(1-2x),3x,3x) = \frac{4}{3(1-x)(4-9x+18x^2)}+\frac{1}{3x(4-9x^2)}$$ The graph of this function is seen on the right. As expected, there is a positive singularity at $x=0$ . Then the function decreases uniformly until it reaches its minimum $f(1/3)=1$ at the center of the triangle. After that it raises a little bit until that local maximum $f(1/2)=22/21$ in the middle of an edge. That $f(1/3)=1$ is indeed a minimum can be verified by calculating $f'(1/3) = 0$ .

• A local minimum. There is something that is not completely clear to me: how do you exclude the existence of other stationary points (minima, indeed) inside the triangle (like $(\alpha,\beta,\gamma)$ and cyclic permutations) where the value of the function may be below one? – Jack D'Aurizio Jan 11 '14 at 14:30

It is possible to prove a slightly weaker bound by cutting down the dimension of the problem.

For any $K\in[0,3]$, define: $$f_K(x)=\frac{2}{(3-x)(4-x(K-x))},\qquad g_K(x)=f_{K}(x)+f_{K}(K-x).$$ By differentiating with respect to $x$, we have that the minima of $g_K(x)$ over $[0,3]$ are located in $\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}$, so $$g_K(x) \geq g_K\left(\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}\right)=\frac{8(6-K)}{(13-3K)^2}$$ holds over $[0,K]$. By taking $x=b,K=b+c$, $a=3-K$ follows and we have: $$\sum_{cyc}\frac{2}{(3-a)(4-bc)}=\frac{2}{K(4-x(K-x))}+g_K(x).\tag{1}$$ The first term in the RHS is a non-negative and convex function $h_K(x)$ on $[0,K]$ whose graphics is symmetric with respect to $x=K/2$. Since $h_K(K/2)>h_K(0)$, $h_K(x)\geq h_K(0)=\frac{1}{2K}$ follows. This gives: $$\sum_{cyc}\frac{2}{(3-a)(4-bc)}\geq\frac{1}{2K}+\frac{8(6-K)}{(13-3K)^2}=j(K).\tag{2}$$ $j(K)$ is a convex function over $[0,3]$, whose minimum is attained in $x=1.4638\ldots$. This gives:

$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq 0.831262\ldots > \frac{4}{5}.\tag{3}$$ We can improve this to: $$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq h_K\left(\frac{K}{2}-\frac{1}{2}\sqrt{1+(3-K)^2}\right)+\frac{8(6-K)}{(13-3K)^2}=\frac{4(13+9K-2K^2)}{K(13-3K)^2},\tag{4}$$ where the RHS is increasing on $[2,3]$. Since we can assume $K=(b+c)\geq 2$ without loss of generality, we have: $$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq\frac{46}{49}.\tag{5}$$

• Quite nice. But it's possible to prove that $\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq 1$ , as requested: see my answer. – Han de Bruijn Jan 10 '14 at 22:04
• @HandeBruijn I don’t think your answer actually constitutes a proof because the Purkiss principle only gives a local minimum. – Ewan Delanoy Jan 11 '14 at 14:25
• @EwanDelanoy: How local is "local". Isn't the inside of a triangle local enough, after that elaborate visualization exercise? (Oh well, I'll take a further look at it and see if an update is necessary) – Han de Bruijn Jan 11 '14 at 14:36

By C-S $$\sum_{cyc}\frac{1}{(a+b)(4-ab)}=\sum_{cyc}\frac{(c+6)^2}{(a+b)(4-ab)(c+6)^2}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(c+6)\right)^2}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}=\frac{441}{\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2}.$$ Thus, it remains to prove that $$882\geq\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2.$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, it's obvious that the last inequality is a linear inequality of $w^3$,

which says that it's enough to prove the last inequality for the extremal value of $w^3$,

which happens in te following cases.

1. $w^3\rightarrow0^+$.

Let $c\rightarrow0^+$ and $b=3-a$.

After this substitution we obtain: $$882-\sum\limits_{cyc}(a+b)(4-ab)(c+6)^2=18>0;$$

1. $b=a$ and $c=3-2a$.

In this case we get $(a-1)^2(2a^3-3a^2+6a+3)\geq0$, which is obvious.

Done!