# Spring 2018, problem 65

Let $a,b,c$ be sides of a triangle. Prove that
$$ 2 < \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}\leq 3 \$$

### Comments

Nice answer, Luciano ! I suppressed my comment for the editors concerning the TeX error (in fact your comment contains the exact formula) On the other hand I see that the rendering of "strictly less" poses a problem also to you. If you find a solution, please post it.

Thanks for the detailed answer.

I think we want to prove that $$ 2 \< \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} - \frac{a^3+b^3+c^3}{abc}\leq 3, $$ which is equivalent to $$ 2abc+a^3 +b^3+c^3 \< ab(a+b)+ac(a+c)+bc(b+c) \leq 3abc+a^3+b^3+c^3. $$ Let's analyze the two inequalities separately.

$1.$

The right inequality is well-known ( $ a^{2}b+ab^{2} +\dots+b^{2}c+bc^{2} \leq3abc+a^3+b^3+c^3 $ ), and results from Schur's Inequality. (see: Art of Problem Solving, Inequalities).

$2.$

In the left inequality, put $a=x+y$, $b=x+z$, $c=y+z$, $a+b=2x+y+z$, and so on, with $x$, $y$ and $z$ positive numbers (to ensure the trianglular inequality): $$2(x+y)(x+z)(y+z)+(x+y)^3+(x+z)^3+(y+z)^3 \< (x+y)(x+z)(2x+y+z)+\dots+ (x+z)(y+z)(2z+x+y).$$ And simple calculation prove the left inequality.