trong bất đẳng thức:(a+b+c)(1/a+1/b+1/c)\geqslant9thay $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)\geqslant9thay
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