An Inequality in Triangle

For $a,b,c\gt 0$ the sides of a triangle

$W(a,b,c)=a^3b+b^3c+c^3a-a^2b^2-b^2c^2-c^2a^2\ge 0$

and $W(a, b, c) = 0$ if and only if $a = b = c.$


Since $a,$ $b$ and $c$ are sides of a triangle, we have $a + b\ge c,$ $b + c \ge a$ and $c + a\ge b.$ Introducing change of variables $a = x + y,$ $b = y + z,$ $c = z + x,$ using that $x, y,z\ge 0,$ and substituting in the equation, we obtain


The main tool in the proof will be the weighted arithmetic mean-geometric means inequality:

$\displaystyle \bigg(\frac{w_{1}F_{1}+w_{2}F_{2}+w_{3}F_{3}}{w_{1}+w_{2}+w_{3}}\bigg)^{w}\ge F^{w_{1}}_{1}F^{w_{2}}_{2}F^{w_{3}}_{3}.$

where $w=w_{1}+w_{2}+w_{3}$ and equality occurs only if $F_{1}=F_{2}=F_{3},$ for positive weights.

The first step is to establish

$\displaystyle \frac{w_{1}x^3y+w_{2}y^3z+w_{3}z^3x}{w_{1}+w_{2}+w_{3}}\ge x^2yz$

using appropriate weights:

$\begin{cases} 3w_{1}+w_{3}&=2w_{1}+2w_{2}+2w_{3}\\ w_{1}+3w_{2}&=w_{1}+w_{2}+w_{3}\\ w_{2}+3w_{3}&=w_{1}+w_{2}+w_{3}. \end{cases}$

Hence, for $w_{1}=4,$ $w_{2}=1,$ $w_{3}=2,$ we obtain

$\displaystyle \frac{4}{7}x^3y+\frac{1}{7}y^3z+\frac{2}{7}z^3x\ge x^2yz.$

Permuting the variables twice, we obtain two more inequalities. Adding all three inequalities, we obtain the desired one. The equality occurs only if $x^3 y = y^3z = z^3x,$ which is equivalent to $x = y = z,$ provided $xyz\ne 0.$ Note that if at least one variable vanishes, say $x = 0,$ and equality occurs, then $y^3z = 0$ and either $a = x + y = 0$ or $c = z + x = 0,$ contradicting our assumption.


The inequality above has been published by Maxim Arnold and Vadim Zharnitsky in The American Mathematical Monthly (April 2015, 378-379) as an ancillary lemma in a bigger article. Arnold and Zharnitsky found the inequality useful. I lifted it from their article, being charmed with the proof.

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