Bài 103209

Question

Cho $a,b,c,d>0$.Chứng minh rằng:
$\frac{a^{2}}{b^{5}}+\frac{b^{2}}{c^{5}}+\frac{c^{2}}{d^{5}}+\frac{d^{2}}{a^{5}}\geq \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}}$

score 0 · Answer 1

Theo BĐT Cauchy:
$\frac{a^{2}}{b^{5}}+\frac{a^{2}}{b^{5}}+\frac{a^{2}}{b^{5}}+\frac{1}{a^{3}}+\frac{1}{a^{3}}\geq 5\sqrt[5]{\left ( \frac{a^{2}}{b^{5}}\right )^{3}\left ( \frac{1}{a^{3}}\right )^{2}}$
$\Rightarrow \frac{3a^{2}}{b^{5}}+\frac{2}{a^{3}}\geq \frac{5}{b^{3}}$ $\left ( 1 \right )$
Tương tự ta có:
$ \frac{3b^{2}}{c^{5}}+\frac{2}{b^{3}}\geq \frac{5}{c^{3}}$ $\left ( 2 \right )$
$ \frac{3c^{2}}{d^{5}}+\frac{2}{c^{3}}\geq \frac{5}{d^{3}}$ $\left ( 3 \right )$
$ \frac{3d^{2}}{a^{5}}+\frac{2}{d^{3}}\geq \frac{5}{a^{3}}$ $\left ( 4 \right )$
Cộng $\left ( 1 \right )$,$\left ( 2 \right )$,$\left ( 3 \right )$,$\left ( 4 \right )$ vế với vế,ta được:
$3\left ( \frac{a^{2}}{b^{5}}+\frac{b^{2}}{c^{5}}+\frac{c^{2}}{d^{5}}+\frac{d^{2}}{a^{5}} \right )\geq3\left ( \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}+\frac{1}{d^{3}} \right )$
$\Rightarrow $ (ĐPCM)

Bài 103209

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Bài 103209

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