work 011 – Kvant No.4 1990

Problem

Prove the following inequality –

$ax + by +cz + \sqrt{(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)} \ge \frac{2}{3} \cdot A$

where

$A = (a+b+c)(x+y+z)$

for Reals $a,b,c,x,y,z$

My Solution

The inequality is cyclic. So assume that $b,y$ are the middle elements of $(a,b,c) ; (x,y,z)$ respectively.

By Cauchy-Schwarz Inequality,

$\sqrt{(a^2 + b^2 + c^2)(x^2 + y^2 + z^2)} \ge az + by + cx$

Therefore it is sufficient to prove that –

$(a + c)(x + z) + 2by \ge RHS$

$\Leftrightarrow (x + z - 2y)(a + c - 2b) \ge 0$

which is true from the assumption .

Hence the inequality is proved with equality for $a=b=c$ and $x=y=z$