Bài 1: Cho a,b,c>0
Chứng minh $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\geq ab+bc+ca$
Áp dụng BĐT Cô-si (AM-GM), ta có :
$\frac{a^3}{b}+\frac{b^3}{c} + bc \geq 3ab$
$\frac{b^3}{c}+\frac{c^3}{a} + ac \geq 3bc$
$\frac{a^3}{b}+\frac{c^3}{a} + ab \geq 3ac$
$\Rightarrow \frac{a^3}{b}+\frac{b^3}{c} + bc + \frac{b^3}{c}+\frac{c^3}{a} + ac + \frac{a^3}{b}+\frac{c^3}{a} + ab \geq 3(ab + bc + ac)$
$\Rightarrow 2.(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}) + ab + bc + ac \geq 3(ab + bc + ac)$
$\Rightarrow 2.(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}) \geq 2(ab + bc + ac)$
$\Rightarrow \frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab+bc+ca$
Dấu "=" xảy ra $\Leftrightarrow a = b = c$
Bài viết đã được chỉnh sửa nội dung bởi Silverbullet069: 14-08-2015 - 13:56
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Unknown to Death, Nor known to Life,
So as I pray, unlimited blade works."