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trong bất đẳng thức$:(a+b+c)(1/a+1/b+1/c)\geqslant9$thay $a=(x+y;b=x+z;c=x+y,ta đc: $$\Leftrightarrow 2(x+y+z)(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\geq9$$ \Leftrightarrow (x+y+z)$________________________________$\geq 9/2 $$\Leftrightarrow \frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}+\frac{x+y+z}{x+y}\geq 9/2$$ \Leftrightarrow \frac{x}{y+z}+1+\frac{y}{x+z}+1+\frac{z}{x+y}+1\geq9/2$$\Rightarrow đpcm$

trong bất đẳng thức$:(a+b+c)(1/a+1/b+1/c)\geqslant9$thay $a=x+y;b=x+z;c=x+y,ta đc:$$\Leftrightarrow 2(x+y+z)(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\geq9$$\Leftrightarrow (x+y+z)$________________________________$\geq 9/2$$\Leftrightarrow \frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}+\frac{x+y+z}{x+y}\geq 9/2$$\Leftrightarrow \frac{x}{y+z}+1+\frac{y}{x+z}+1+\frac{z}{x+y}+1\geq9/2$$\Rightarrow đpcm$