a,b,c>0, abc=1. CMR:
b, $2(a^{2}+b^{2}+c^{2})+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3(a+b+c)$
b) Ta đã c/m được : $a^{2}+b^{2}+c^{2} + \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2(a + b + c)$
Giờ cần phải c/m : $a^{2}+b^{2}+c^{2} \geq a + b + c$
Ta có : $a^{2}+b^{2}+c^{2} = \frac{a^2}{1} + \frac{b^2}{1} + \frac{c^2}{1} \geq \frac{(a + b + c)^2}{3}$
Áp dụng BĐT Cô-si, ta có :
$a + b + c \geq 3.\sqrt[3]{abc} = 3.\sqrt[3]{1} = 3$
$\Rightarrow \frac{(a + b + c)^2}{3} = \frac{(a + b + c).(a + b + c)}{3} \geq \frac{3(a + b + c)}{3} = a + b + c$
$\Rightarrow a^{2}+b^{2}+c^{2} \geq a + b + c$
Vậy $2(a^{2}+b^{2}+c^{2})+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 3(a+b+c)$
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Unknown to Death, Nor known to Life,
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