[Toán 10] chứng minh BDT

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0915549009

Cm Cho a,b >0
1.[TEX]\frac{1}{a} +\frac{1}{b} + \frac{1}{c} \ge 3[/TEX] với a+b+c=3
2. [TEX](1+ \frac{1}{a}+b)^3 + (1+\frac{1}{b}+a)^3 \geq 16[/TEX]
[TEX]1)\frac{1}{a} + \frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}=3[/TEX]
[TEX]2)(1+\frac{1}{a}+b)^3+(1+\frac{1}{b}+a)^3 \geq (2+\frac{1}{a}+a+\frac{1}{b}+b)(1+\frac{1}{a}+b)(1+\frac{1}{b}+a) \geq6.9=54[/TEX] :-??
 
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