[Đại8]Bất đẳng thức khó

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1,BĐT \Leftrightarrow $x+y+z-xyz \le 2$

\Leftrightarrow $x(1-yz)+y+z \le 2$

Ta có: $[x(1-yz)+y+z]^2 \le (x^2+y^2+z^2+2yz)(1-2yz+y^2z^2+1)=2(1+yz)(y^2z^2-2yz+2)$

Cần cm $2(1+yz)(y^2z^2-2yz+2) \le 4$

\Leftrightarrow $y^2z^2(yz-1) \le 0$

Có $yz \le \dfrac{y^2+z^2}{2} \le 1$

\Rightarrow đpcm.

2,$(x+y)(\dfrac{2}{x}+\dfrac{3}{y}) \ge (\sqrt{x}.\sqrt{\dfrac{2}{x}}+\sqrt{y}.\sqrt{\dfrac{3}{y}})^2=(\sqrt{2}+\sqrt{3})^2$

\Leftrightarrow $x+y \ge \dfrac{5+2\sqrt{6}}{6}$
 
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