[warutT]

ข้อ 33 ครับ

From Cauchy-Schwarz Inequality

$ab+bc+ca \leq a^2+b^2+c^2 ...(*)$

$2(ab+bc+ca) \leq 2(a^2+b^2+c^2)$

$a^2+b^2+c^2+2(ab+bc+ca) \leq 3(a^2+b^2+c^2)$

$(a+b+c)^2 \leq 3(a^2+b^2+c^2) ...(**)$

From modified Cauchy-Schwarz Inequality

$\dfrac{a^2}{b^2+bc+c^2}+\dfrac{b^2}{c^2+ca+a^2}+\dfrac{c^2}{a^2+ab+b^2}$

$\geq \dfrac{(a+b+c)^2}{b^2+bc+c^2+c^2+ca+a^2+a^2+ab+b^2}$

$=\dfrac{(a+b+c)^2}{2(a^2+b^2+c^2)+ab+bc+ca}$

$\geq \dfrac{(a+b+c)^2}{2(a^2+b^2+c^2)+a^2+b^2+c^2} (From (*))$

$=\dfrac{(a+b+c)^2}{3(a^2+b^2+c^2)}$

$\geq \dfrac{(a+b+c)^2}{(a+b+c)^2}=1 (From (**))$

$\therefore \dfrac{a^2}{b^2+bc+c^2}+\dfrac{b^2}{c^2+ca+a^2}+\dfrac{c^2}{a^2+ab+b^2}\geq 1$

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