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} .....

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