Consider the following quantities for $a, b, c \in \mathbb{R}^{+}$ and $a\neq b$.

$$\begin{align} &x_1 = 3 \left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a}\right)^{-1} \\ &x_2 = \frac{ab + bc + ac}{a^2 + b^2 + c^2} \\ &x_3 = \frac{3}{2}\left( \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b}\right)^{-1}\end{align}$$

Is there a simple way to know which one is the largest and check if it is less than $1$?

  • $\begingroup$ What kind of "simple way" are you expecting? It can be easily proven that $x_1< 1$ always, $x_2< 1$ always, and $x_2 < x_3$ always (while $x_3$ itself can be above or beyond $1$). The boundary surface defined by $x_1 = x_3$ allows an expression as an implicit function of two variables like $F(t, \theta) = 0$. However, to make it into an explicit function (like an ordinary $z = f(x,y)$ single valued surface) is to deal with a quartic polynomial which in this case doesn't permit an analytic solution. You might as well do it numerically to begin with. $\endgroup$ May 25, 2018 at 9:34
  • $\begingroup$ @LeeDavidChungLin Can you show me, how u did x2< x1 also x2<x3. $\endgroup$
    – Tigist
    May 25, 2018 at 18:09
  • $\begingroup$ I never said anything that implies $x_2 < x_1$. However, I do have to apologize for the mistake in my previous comment: $x_3$ cannot go beyond $1$. That is, all three of your quantities are less than $1$ always. $\endgroup$ May 25, 2018 at 22:26

2 Answers 2


After this intro will come first the proof for $x_2 < x_3$ that is really elementary (albeit seemingly long on the screen). Then, a brief discussion of the relation between $x_1$ and $x_3$.

I will not address the easy proofs of $x_1 < 1$ and $x_2 < 1$, assuming you know them.

The proof that $x_3 < 1$ always can be done (among many ways) by a change of variables. BTW, this is known as the Nesbitt's inequality and one can also see this. $$ x_3 \overset{?}{<} 1 \quad \Longleftrightarrow \quad 2 \left(\frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) \overset{?}{>} 3 \\ \left\{ \begin{aligned} \alpha &\equiv \frac{b + c}2 \\ \beta &\equiv \frac{c + a}2 \\ \gamma &\equiv \frac{a + b}2 \end{aligned} \right. \quad \implies \left\{ \begin{aligned} a &= \beta + \gamma - \alpha \\ b &= \gamma + \alpha - \beta \\ c &= \alpha + \beta - \gamma \end{aligned} \right. \quad \implies \text{L.H.S.} = \frac{\beta + \gamma - \alpha}{\alpha} + \frac{\gamma + \alpha - \beta}{\beta} + \frac{\alpha + \beta - \gamma }{\gamma } $$ Note that each term gives a $-1.~$ Rearrange and pair terms up as $$\text{L.H.S.} = \left( \frac{\beta}{\alpha} + \frac{\alpha}{\beta} \right) + \left( \frac{\gamma}{\beta} + \frac{\beta}{\gamma} \right) + \left( \frac{\alpha}{\gamma} + \frac{\gamma}{\alpha} \right) - 3$$ Each parenthesis is at least two due to AM-GM inequality. For example, $\frac{\beta}{\alpha} + \frac{\alpha}{\beta} \geq 2\sqrt{ \frac{\beta}{\alpha} \frac{\alpha}{\beta}} = 2$.

Therefore, we have $\text{L.H.S.} \geq 6 - 3 = 3$, which proves $x_3 > 1$. The equality occurs only when $a = b = c$, which is excluded by definition that $a \neq b$.$\quad \mathbf{Q.E.D.}$

Proof of $x_2 < x_3$

I will arrange the terms into multi-lines within a big parenthesis to emphasize the cyclic pattern. These shall not be confused with matrices.

The $\Longleftrightarrow$ below are just elementary operations (multiplying and rearranging the terms). The inequality is not "pushed" and it remains equivalent all the way up to \ref{Eq_01}.

In the first line, we can cross multiply the denominators because $\{a,b,c\}$ are all positive and so are their combo.

\begin{align*} x_2 &\overset{?}{<} x_3 &&\Longleftrightarrow & 2(ab + bc + ac)\left( \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) &\overset{?}{<} 3(a^2 + b^2 + c^2) \\ & &&\Longleftrightarrow & \left( \begin{aligned}[c] ~ab &+ bc \\ &+bc+ ac \\ {}+ab &\qquad\!+ac\end{aligned} \right)\left( \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) &\overset{?}{<} 3(a^2 + b^2 + c^2)\\ & &&\Longleftrightarrow & \left( \begin{aligned}[c] &b(a + c) \\ &\quad +c(a + b) \\ &\qquad +a(b+c)\end{aligned} \right)\left( \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) &\overset{?}{<} 3(a^2 + b^2 + c^2) \\ & &&\Longleftrightarrow & \begin{pmatrix} ab\dfrac{a+c}{\color{magenta}{b + c}} &+ & b^2 &+ & bc\dfrac{a+c}{\color{blue}{a + b} } \\ +ac\dfrac{a+b}{\color{magenta}{b + c}} &+ &bc\dfrac{a+b}{a + c} &+ &c^2 \\ +a^2 &+ &ab\dfrac{b+c}{a + c} &+ &ac\dfrac{b+c}{ \color{blue}{a + b} }\end{pmatrix} &\overset{?}{<} 3(a^2 + b^2 + c^2) \end{align*} Here, one set of $a^2+b^2+c^2$ cancels. We continue by gathering the terms on the L.H.S. with the common denominator. For example, those in magenta go together and those in blue go together. \begin{align*} x_2 &\overset{?}{<} x_3 &&\Longleftrightarrow & \begin{pmatrix} \dfrac1{\color{magenta}{b + c}}\bigl( ab(a+c) + ac(a+b)\bigr) \qquad \\ \quad + \dfrac1{a+c}\bigl( bc(a+b) + ab(b+c)\bigr) \\ \qquad +\dfrac1{\color{blue}{a+b}}\bigl( bc(a+c) + ac(b+c)\bigr) \end{pmatrix} &\overset{?}{<} 2(a^2 + b^2 + c^2) \\ & &&\Longleftrightarrow & \begin{pmatrix} \dfrac1{b + c}\bigl( a^2(b+c) + 2abc\bigr) \qquad \\ \quad + \dfrac1{a+c}\bigl( b^2(a+c) + 2abc \bigr) \\ \qquad +\dfrac1{a+b}\bigl( c^2(a+b) + 2abc \bigr) \end{pmatrix} &\overset{?}{<} 2(a^2 + b^2 + c^2) \end{align*} At this point, another set of $a^2+b^2+c^2$ cancels and we have \begin{align*} x_2 &\overset{?}{<} x_3 &&\Longleftrightarrow & 2abc \bigl( \frac1{b + c} + \frac1{a + c} + \frac1{a + c} \bigr) &\overset{?}{<} a^2 + b^2 + c^2 \tag*{Eq.(1)} \label{Eq_01} \end{align*} Again, the $\Longleftrightarrow$ so far are just multiplying and rearranging the terms. Here's where we start pushing the inequality.

Since $b,c >0$ we have the AM-GM inequality: $~\sqrt{bc} \leq \dfrac{b+c}2 \implies 2bc \leq \dfrac{ (b + c)^2 }2$. This renders the first term on the left hand side of \ref{Eq_01} to be: $~a\dfrac{2bc}{b + c} \leq a \dfrac{b+c}2$. Do the same the other two terms and push the L.H.S. of \ref{Eq_01} as $$2abc \bigl( \frac1{b + c} + \frac1{a + c} + \frac1{a + c} \bigr) \leq ab + bc + ac $$ Finally, with the basic $ab \leq \dfrac{a^2 + b^2}2$ which comes from $(a - b)^2 \geq 0$, we have $$ ab + bc + ac \leq \frac{a^2 + b^2}2 + \frac{b^2 + c^2}2 + \frac{c^2 + a^2}2 = a^2 + b^2 + c^2 $$ which proves \ref{Eq_01}. The equality holds only when $a = b = c$, where $x_2 = x_3 = 1$, which never happens because the specified $a \neq b$.$\quad \mathbf{Q.E.D.}$

The above proves $x_2 < x_3$ always, hence $\max\{x_1,\, x_2,\, x_3\}$, can only be $x_1$ or $x_3$. Namely, we can ignore $x_2$ completely when considering largest of the three.

Surface that separates $x_1 > x_3$ and $x_1 < x_3$

The region where $x_1 \geq x_3$ is the 3 disconnected "branches" enclosed inside the yellow surfaces in the figure below. That is, we have $x_3$ being the largest in most of the space.

The blue surfaces are "fake" and are merely there to show the cutoff at the box (plotting range). In the plot the axes $X$, $Y$, and $Z$ stands for $a$, $b$ and $c$, respectively.

enter image description here

The right figure is the top view staring straight down into the branch where $c > a > b$. Note that there are a total of six orders for the three things $(a,b,c)$, but the boundary of $x_1 = x_3$ for this branch doesn't go into $c > b > a$ and is completely inside $c > a > b$.

That is, despite $x_1$ and $x_3$ individually being symmetric in all six permutations of $(a,b,c)$, there are only three branches and not six. The three branches are cyclic: they are 3-fold-rotationally invariant ($a \mapsto b,\, b \mapsto c,\, c \mapsto a$) but not swapping invariant ($ a \leftrightarrow b$).

Also note that the surface is "intercepted" at $b = 0$ (the $X$-$Z$ plane at $Y = 0$).

The fact that the three branches are disconnected (except at $(0,0,0)$ the origin) means one can consider the boundary surface $x_1 = x_3$ separately for each branch.

Again, take the $Z$-branch in $c > a > b$ in the right figure ($Z > X > Y$) for example. The surface $x_1 = x_3$ as the implicit function is $$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 2 \left( \frac{a}{b + c} + \frac{b}{a + c} + \frac{c}{a + b} \right) \tag*{Eq.(2)} \label{Eq_02}$$ Despite the pretty functional form with nice symmetry, to solve for the explicit $c = f(a,b)$, namely, $z = f(x,y)$, is to solve the roots of a quartic polynomial (4-th power in $z$).

Quartic equations indeed permits analytic solutions in general (the roots can be expressed in radicals). Unfortunately, the surface of each branch consists of two pieces with a critical (split) point that allows no analytic expression. Namely, for the $Z$-branch, given a position $(x,y)$ one can obtain the height $z$ (the corresponding point on the surface) only after numerically testing which piece this point belongs to via checking $x \gtrless x_0$, where $x_0$ is a root of a $16$-th degree equation. (the degree of $16 = \text{4-by-4}$ is not a coincidence)

The implicit function \ref{Eq_02} of three variables can be rewritten into (another implicit function) $$\cot\phi + \frac{t}{ \cos\phi } + \frac{ \sin\phi }{ t } = 2\left( \frac{ \sin\phi }{ t + \cos\phi } + \frac{ \cos\phi }{ t + \sin\phi } + \frac{ t }{ \cos\phi + \cos\phi } \right) \tag*{Eq.(3)} \label{Eq_03}$$ where $\phi \equiv \arctan( \dfrac{x}{y} )$ is the azimuthal angle and $t \equiv \dfrac{z}{ \sqrt{x^2 + y^2} }$ is the cylindrical tangent.

There are many other re-parametrizations that either make the equation more compact (involving only 1 variable) or more geometric (invoking $x>y \Longleftrightarrow 0< \phi < \pi/4$ and introduce more trigs). As far as I can tell, none of them alters the fact that the surface is quartic in nature. As a whole, \ref{Eq_03} has a better balance among the many equivalent expressions.

  • $\begingroup$ @Tigist Glad I could help. Perhaps you can consider clicking the accept-button (the green 'check') to "conclude" this post. Otherwise, it hangs loose in the queue and clogs the system a little bit. Thank you. $\endgroup$ May 27, 2018 at 1:55
  • $\begingroup$ oh I thought I clicked the green one. Have a nice day:) $\endgroup$
    – Tigist
    May 27, 2018 at 3:06

By AM-GM $$x_1=\frac{3}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}\leq\frac{3}{3\sqrt[3]{\prod\limits_{cyc}\frac{a}{b}}}=1;$$ $$x_2=\frac{ab+ac+bc}{a^2+b^2+c^2}=1-\frac{\sum\limits_{cyc}(a^2-ab)}{a^2+b^2+c^2}=$$ $$=1-\frac{\sum\limits_{cyc}(a^2-2ab+b^2)}{2(a^2+b^2+c^2)}=1-\frac{\sum\limits_{cyc}(2a^2-2ab)}{2(a^2+b^2+c^2)}=1-\frac{\sum\limits_{cyc}(a-b)^2}{2(a^2+b^2+c^2)}\leq1 $$ and by C-S $$x_3=\frac{\frac{3}{2}}{\sum\limits_{cyc}\frac{a}{b+c}}=\frac{\frac{3}{2}}{\sum\limits_{cyc}\frac{a^2}{ab+ac}}\leq\frac{\frac{3}{2}}{\frac{(a+b+c)^2}{\sum\limits_{cyc}(ab+ac)}}=\frac{3(ab+ac+bc)}{(a+b+c)^2}\leq1,$$ where the last inequality it's just $x_2\leq1,$ which is proven.

Now, we'll prove that $x_2\leq x_3$.

Indeed, we need to prove that $$\frac{ab+ac+bc}{a^2+b^2+c^2}\leq\frac{\frac{3}{2}}{\sum\limits_{cyc}\frac{a}{b+c}}$$ or $$\sum\limits_{cyc}\frac{a}{b+c}\leq\frac{3(a^2+b^2+c^2)}{2(ab+ac+bc)}$$ or $$\sum\limits_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)\leq\frac{3(a^2+b^2+c^2)}{2(ab+ac+bc)}-\frac{3}{2}$$ or $$\sum_{cyc}\frac{a-b-(c-a))}{b+c}\leq\frac{\sum\limits_{cyc}(a^2-ab)}{ab+ac+bc}$$ or $$\sum_{cyc}(a-b)\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\leq\frac{3\sum\limits_{cyc}(a^2-2ab+b^2)}{2(ab+ac+bc)}$$ or $$\sum_{cyc}(a-b)^2\left(\frac{3}{2(ab+ac+bc)}-\frac{1}{(a+c)(b+c)}\right)\geq0$$ or $$\sum_{cyc}\frac{(a-b)^2(3c^2+ab+ac+bc)}{(a+c)(b+c)}\geq0,$$ which is obvious.

Now, we'll show that we can't compare $x_1$ and $x_2$.

Indeed, $$x_2-x_1=\frac{ab+ac+bc}{a^2+b^2+c^2}-\frac{3}{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}=\frac{\sum\limits_{cyc}ab\sum\limits_{cyc}a^2c-3\sum\limits_{cyc}a^3bc}{\sum\limits_{cyc}a^2\sum\limits_{cyc}a^2c}=$$ $$=\frac{\sum\limits_{cyc}(a^3c^2+a^3bc+a^2b^2c-3a^3bc)}{\sum\limits_{cyc}a^2\sum\limits_{cyc}a^2c}=\frac{\sum\limits_{cyc}(a^3c^2-2a^3bc+a^2b^2c)}{\sum\limits_{cyc}a^2\sum\limits_{cyc}a^2c}.$$ We see that $x_2>x_1$ for $b\rightarrow0^+$ and $x_2-x_1\rightarrow-\infty$ for $a\rightarrow+\infty$ and $c-2b<0$.


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