# work 012 – VMEO 2006

For $a,b,c$ arbitrary reals . Let –

$x = \sqrt{a^2 - ab + b^2}$

$y = \sqrt{b^2 - bc + c^2}$ ,

$z= \sqrt{c^2 - ca + a^2}$

Prove that the following inequality holds good –

$\sum_{cyclic} xy \ge \sum_{cyclic} a^2$

Proof

note that –

$x = (a -\frac{b}{2})^2 + (\frac{\sqrt{3}b}{2})^2$

$xy = \sqrt{[(a -\frac{b}{2})^2 + (\frac{\sqrt{3}b}{2})^2] \cdot [(c -\frac{b}{2})^2 + (\frac{\sqrt{3}b}{2})^2}]$

By Cauchy-Schwarz Inequality,

$xy \ge (a-\frac{b}{2})(c-\frac{b}{2}) + \frac{3b^2}{4}$

$\sum_{cyclic} xy \ge \sum_{cyclic} a^2$

Hence Proved with equality for $a=b=c$ .

Some results can be proved with the definition – $x = \sqrt{a^2 + ab +b^2} .....$