work 028 – SII Vol.1


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}


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


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