work 028 – SII Vol.1

Problem

Prove that for positive reals $x,y,z$ with sum 1 ,

$\sqrt{x+(y-z)^2}+\sqrt{y+(z-x)^2}+\sqrt{z+(x-y)^2}\ge \sqrt{3}$

Proof

Homogenize the inequality to –

$\sum \sqrt{x(x+y+z)+(y-z)^2} \ge \sqrt{3}\cdot (x+y+z)$

Since the inequality is cyclic, let $z = \min{x,y,z}$

Substitute –

$x \rightarrow x-z$

$y \rightarrow y-z$

$z \rightarrow z-z=0$

and let ,

$x-z = a , y-z = b \implies x-y = a-b$

The inequality by the above substitution is transformed to –

$2\sqrt{a^2+ab+b^2} + |a-b| \ge \sqrt{3}\cdot (a+b)$

By AM-GM inequality ,

$a^2+ab+b^2 \ge \frac{3}{4} \cdot (a+b)^2$

therefore the inequality is true with equality $a=b$

which gives us the equality cases – $x=y=z=\frac{1}{3};x=y=\frac{1}{2},z=0$