Giải thích các bước giải:
1.Ta có :
$A=\dfrac{1}{6}.(6-2y)(12-3x)(2y+3x)$
$\rightarrow A\le \dfrac{1}{6}.(\dfrac{6-2y+12-3x+2y+3x}{3})^3$
$\rightarrow A\le \dfrac{1}{6}.6^3$
$\rightarrow A\le 36$
Dấu = xảy ra khi $6-2y=12-3x=2y+3x$
$\rightarrow x=2,y=0$
2.Ta có :
$\sqrt{z-2}=\dfrac{1}{\sqrt{2}}.\sqrt{(z-2).2}\le \dfrac{1}{\sqrt{2}}.\dfrac{1}{2}.(z-2+2)=\dfrac{1}{2\sqrt{2}}.z$
Chứng minh tương tự ta có :
$\sqrt{x-3}\le \dfrac{1}{2\sqrt{3}}.x$
$\sqrt{y-4}\le \dfrac{1}{2\sqrt{4}}.y$
$\rightarrow \dfrac{xy\sqrt{z-2}+yz\sqrt{x-3}+zx\sqrt{y-4}}{xyz}\le \dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+ \dfrac{1}{2\sqrt{4}}$
$\rightarrow đpcm$