[dektep]

43.(vietnam 2002,Dung Tran Nam)
Solution : Let $x^2\geq y^2 \geq z^2 \Rightarrow x^2 \geq 3, 6\geq y^2+z^2 \geq 2yz$.

cauchy;

$(2(x+y+z)-xyz)^2=(2(y+z)+x(2-yz))^2$

$\leq ((y+z)^2+x^2)(4+(2-yz)^2)=(2yz+9)(y^2z^2-4yz+8)$

Let $\alpha =yz$.

จะพิสูจน์ว่า

$100 \geq (2 \alpha + 9)(\alpha^2-4\alpha +8)$

แต่

$100-(2\alpha +9)(\alpha^2-4\alpha+8)= -2\alpha^3-\alpha^2+20\alpha+28$

$=(\alpha+2)^2(7-2\alpha)\geq 0$