C/m:$$(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})\geq (1+\frac{1}{a})^{3}$$
Ta có $$A=(1+\frac{1}{x})(1+\frac{1}{y})(1+\frac{1}{z})=\frac{xyz+xy+xz+yz+x+y+z+1}{xyz}=\frac{xyz+xy+xz+yz+a+1}{xyz}=\frac{a+1}{xyz}+1+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
Có $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}=\frac{9}{a} (do x,y,z> 0)$$
có $$27xyz\leq (x+y+z)^{3}=a^{3}$$
$$\Rightarrow \frac{a+1}{xyz}\geq \frac{27a+27}{a^{3}}$$
Do đó $$A\geq 1+\frac{9}{a}+\frac{27a}{a^{3}} =\frac{(a+3)^{3}}{a^{3}}> \frac{(a+1)^{3}}{a^{3}}=\left ( 1+\frac{1}{a} \right )^{3}$$
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Bài viết đã được chỉnh sửa nội dung bởi perfectstrong: 08-01-2012 - 22:11
