Question

[Long Vũ]

Câu bất đẳng thức hay

Answer 1

[Lightning Farron]

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

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

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 \(\left\{{}\begin{matrix}x=6\\y=8\\z=4\end{matrix}\right.\)

Answer 2

[Lightning Farron]

hay cái j cha nội ngược dấu kìa :v