inequalities

**the undertaker** · June 24th 2010, 01:18 AM

Prove that

a^4 + b^4 + c^4 > (a^2 * bc) + (b^2 * ac) + (c^2 * ab)

holds for all real numbers a, b, c.

**Opalg** · June 27th 2010, 12:16 PM

Originally Posted by the undertaker

Prove that

$a^4 + b^4 + c^4 > a^2 bc + b^2 ac + c^2 ab$

holds for all real numbers a, b, c.

First note that $0\leqslant (b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2 = 2(a^4 + b^4 + c^4) - 2(b^2c^2+c^2a^2+a^2b^2)$ , from which $b^2c^2+c^2a^2+a^2b^2 \leqslant a^4 + b^4 + c^4$ .

Then by the Cauchy–Schwarz inequality $a^2 bc + b^2 ac + c^2 ab \leqslant (a^4 + b^4 + c^4)^{1/2}(b^2c^2+c^2a^2+a^2b^2)^{1/2} \leqslant a^4 + b^4 + c^4$ .

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