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[alexander123]

Bất đẳng thức: 

 

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[Hoang72]

$\sqrt{\frac{a(b+c)}{a^2+bc}}\leq \frac{1}{2}\left ( 1+\frac{a(b+c)}{a^2+bc} \right )=\frac{(a+b)(a+c)}{2(a^2+bc)}$.

Tương tự suy ra $\sum\sqrt{\frac{a^2+bc}{a(b+c)}}\geq \sum\frac{2(a^2+bc)}{(a+b)(a+c)}=\frac{4\sum_{sym}a^2b}{(a+b)(b+c)(c+a)}=4-\frac{8abc}{(a+b)(b+c)(c+a)}\geq 4-\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}$.

Ta có đpcm.