Bài 102871

Question

Cho $a,b,c,d>0.$Chứng minh rằng:
$\frac{a}{b+c}+ \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \geq 2$

score 0 · Answer 1

Áp dụng BĐT BCS::
$\left ( a+b+c+d \right )^{2}=$
$\left(\sqrt{\frac{a}{b+c}} \sqrt{a\left (b+c \right )} + \sqrt{\frac{b}{c+d}} \sqrt{b\left ( c+d\right )}+ \sqrt{\frac{c}{d+a}} \sqrt{c\left ( d+a\right )}+\sqrt{\frac{d}{a+b}} \sqrt{d\left (a+b \right )}\right )^{2}$
$ \leq \left ( \frac{a}{b+c}+ \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \right )\left ( ab+ac+bc+bd+cd+ca+da+db \right ) $
$\Rightarrow \frac{a}{b+c}+ \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b} \geq \frac{\left ( a+b+c+d \right )^{2}}{ ab+ac+bc+bd+cd+ca+da+db }$
Ta sẽ chứng minh:
$\frac{\left ( a+b+c+d \right )^{2}}{ ab+ac+bc+bd+cd+ca+da+db } \geq 2$
$\Leftrightarrow a^{2}+b^{2}+c^{2}+d^{2}+2 \left ( ab+ac+ad+bc+bd+cd \right ) \geq $
$\geq 2 \left ( ab+ac+bc+bd+cd+ca+da+db \right ) $
$\Leftrightarrow a^{2}+b^{2}+c^{2}+d^{2} -2ac-2bd\geq 0\Leftrightarrow \left ( a-c \right )^{2}+ \left ( b-d \right )^{2}\geq 0 $ (luôn đúng)
$\Rightarrow $ (ĐPCM)

Bài 102871

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Bài 102871

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