Làm giúp mình vs càn gấp mai thi r

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linh123658

$\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$
Áp dụng AM-GM cho các số không âm$\dfrac{a^2}{b^2};\dfrac{b^2}{c^2};\dfrac{c^2}{a^2}$
Ta có:$\frac{a^2}{b^2}+\frac{b^2}{c^2}$\geq$2\dfrac{a}{c}$
$\dfrac{a^2}{b^2}+\dfrac{c^2}{a^2}$\geq$2\dfrac{c}{b}$
$\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}$\geq$2\dfrac{b}{a}$
Cộng vế với vế ra $đpcm$
 
X

xuanquynh97

Theo bất đẳng thức Cauchy ta có
$\frac{a^2}{b^2}+\frac{b^2}{c^2}$ \geq $2\frac{a}{c}$
$\frac{b^2}{c^2}+\frac{c^2}{a^2}$\geq $\frac{b}{a}$
$\frac{c^2}{a^2}+\frac{a^2}{b^2}$\geq $\frac{c}{b}$
\Rightarrow $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}$ \geq $\frac{b}{a}+\frac{a}{c}+\frac{c}{b}$
Dấu = xảy ra khi $\frac{a}{b}$ = $\frac{b}{c}$ = $\frac{c}{a}$
 
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