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[Sagittarius912]

Cho $a,b,c>0$ thỏa mãn $a+b+c=3$. Chứng minh:

$\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3$

[Nguyen Duc Thuan]

Cho $a,b,c>0$ thỏa mãn $a+b+c=3$. Chứng minh:

$T=\sqrt{\frac{a+b}{c+ab}}+\sqrt{\frac{b+c}{a+bc}}+\sqrt{\frac{c+a}{b+ca}}\ge 3$

Áp dụg AM-GM:

$T\geq 3\sqrt[6]{\prod \frac{a+b}{c+ab}}$ (*)

MÀ: $(ab+c)(bc+a)\leq \left ( \frac{a+c+ab+bc}{2} \right )^2=\left ( \frac{(a+c)(b+1)}{2} \right )^2$

$\Rightarrow \prod (ab+c)^2\leq \frac{\prod (a+b)^2\prod (a+1)^2}{64}\leq (a+b)^2$ (CAUCHY Cho $\sum (a+1)$)

$\Rightarrow \frac{\prod (a+b)}{\prod (c+ab)}\geq 1$

Thay vào (*) có đpcm !