[warutT]

ข้อ 30 ครับ

From Cauchy-Schwarz Inequality

$ac+bd \leq \sqrt{a^2+c^2}\sqrt{b^2+d^2}$

From A.M.-G.M. Inequality

$\sqrt{a^2+c^2}\sqrt{b^2+d^2} \leq \dfrac{a^2+b^2+c^2+d^2}{2}$

$\therefore ac+bd \leq \dfrac{a^2+b^2+c^2+d^2}{2}$

$a^2+b^2+c^2+d^2 \geq 2(ac+bd)$

$a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd \geq 2ab+2ac+2ad+2bc+2bd+2cd+2(ac+bd)$

$(a+b+c+d)^2 \geq 2ab+2ac+2ad+2bc+2bd+2cd+2ac+2bd$

$\dfrac{(a+b+c+d)^2}{2} \geq ab+ac+ad+bc+bd+cd+ac+bd$

$\dfrac{1}{ab+ac+ad+bc+bd+cd+ac+bd} \geq \dfrac{1}{\dfrac{(a+b+c+d)^2}{2}} ...(*)$

From modified Cauchy-Schwarz Inequality

$\dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}$

$=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ca}+\dfrac{d^2}{da+db}$

$\geq \dfrac{(a+b+c+d)^2}{ab+ac+bc+bd+cd+ca+da+db}$

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

$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+d}+\dfrac{c}{d+a}+\dfrac{d}{a+b}\geq 2$

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