# A small extension of Schur’s Inequality

Of the numerous extension of the Schur’s Inequality , this is a particular result –

For 6 non-negative reals – a,b,c,x,y,z  such that  $\sqrt{x}+\sqrt{z} \ge y$ and $a\ge b\ge c$, we have –

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

Proof

Since $a\ge b\ge c$ , $(a-b)(b-c) \ge 0$.

Therefore,dividing the LHS by $(a-b)(b-c)$,

$x \cdot \frac{a-c}{b-c} + (-y) + z\cdot \frac{a-c}{a-b} \ge 0$

$\Leftrightarrow x \cdot \frac{a-c}{b-c} + z\cdot \frac{a-c}{a-b} \ge y$

$\Leftrightarrow x \cdot \frac{a-b}{b-c} + z\cdot \frac{b-c}{a-b} +x + z \ge y$

$\Leftrightarrow x+z+2\sqrt{xz} \ge y$

$\Leftrightarrow \sqrt{x}+\sqrt{z} \ge \sqrt{y}$

which is true from the condition given. This completes the proof with equality for $a=b=c , x=z=\frac{y}{2}$