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}

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