Seriously, totally, absolutely, definitely not about the Schwarz inequality...

Algebra Level 3

$a$ , $b$ and $c$ are three sides of a triangle.

\[\large \begin{array} {} 4\left(\dfrac{1}{a + b} + \dfrac{1}{b + c} + \dfrac{1}{c + a}\right) & \huge \square & \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} + \dfrac{9}{a + b + c} \end{array}\]

Which sign $<$ , $\le$ , $=$ , $\ge$ , or $>$ should be in the box?

Note: Find the strongly inequality/equality sign.

This is part of the set: It's easy, believe me!

<

=

\ge

>

\le

1 solution

Joe Mansley
Feb 20, 2019

$x^{-1}$ is convex. Therefore, by Popoviciu's inequality, $(\frac{a+b+c}{3})^{-1}+\frac{a^{-1}+b^{-1}+c^{-1}}{3} \geq \frac{2}{3}((\frac{a+b}{2} )^{-1}+(\frac{b+c}{2})^{-1}+(\frac{c+a}{2})^{-1})$ . Rearranging gives the desired result. Equality is achieved when the $a=b=c$ , i.e. the triangle is equalateral.

Seriously, totally, absolutely, definitely not about the Schwarz inequality...

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