Câu hỏi của Long Vũ

Question

Câu bất đẳng thức hay

Lightning Farron · Accepted Answer

$VT=\dfrac{xy\sqrt{z-2}+yz\sqrt{x-3}+xz\sqrt{y-4}}{xyz}$

$=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}$

Áp dụng BĐT AM-GM ta có:

$\dfrac{\sqrt{z-2}}{z}=\dfrac{\sqrt{2\left(z-2\right)}}{\sqrt{2}z}\le\dfrac{\dfrac{2+z-2}{2}}{\sqrt{2}z}=\dfrac{1}{2\sqrt{2}}$

$\dfrac{\sqrt{x-3}}{x}=\dfrac{\sqrt{3\left(x-3\right)}}{\sqrt{3}x}\le\dfrac{\dfrac{3+x-3}{2}}{\sqrt{3}x}=\dfrac{1}{2\sqrt{3}}$

$\dfrac{\sqrt{y-4}}{y}=\dfrac{\sqrt{4\left(y-4\right)}}{\sqrt{4}y}\le\dfrac{\dfrac{4+y-4}{2}}{\sqrt{4}y}=\dfrac{1}{2\sqrt{4}}$

Cộng theo vế 3 BĐT trên ta có:

$VT=\dfrac{\sqrt{z-2}}{z}+\dfrac{\sqrt{x-3}}{x}+\dfrac{\sqrt{y-4}}{y}\le\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}+\dfrac{1}{2\sqrt{4}}=VP$

Dấu "=" xảy ra khi $\left\{{}\begin{matrix}x=6\y=8\z=4\end{matrix}\right.$Câu trả lời: $VT=\dfrac{xy\sqrt{z-2}+yz\sqrt{x-3}+xz\sqrt{y-4}}{xyz}$