Lâu rồi mới thấy 1 bài BĐT hay:

BĐT cần CM tương đương:

$\displaystyle{\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geq 3(a^2+b^2+c^2)\Leftrightarrow (a+b+c)(\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a}) \geq 3(a^2+b^2+c^2)}$

$\displaystyle{\Leftrightarrow \frac{a^{2}(a+c)}{b}+\frac{b^{2}(b+a)}{c}+\frac{c^{2}(c+b)}{a}\geq 2(a^{2}+b^{2}+c^{2})}$

Thật vậy, AM-GM:

$\displaystyle{\frac{a^{2}(a+c)}{b}+(a+c)b\geq 2a(a+c)}$

$\displaystyle{\frac{b^{2}(b+a)}{c}+(b+a)c\geq 2b(b+a)}$

$\displaystyle{\frac{c^{2}(c+b)}{a}+(c+b)a\geq 2c(c+b)}$

Cộng vế theo vế ta có: $\displaystyle{\frac{a^{2}(a+c)}{b}+\frac{b^{2}(b+a)}{c}+\frac{c^{2}(c+b)}{a}\geq 2(a^{2}+b^{2}+c^{2})}$

  (ĐPCM)

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