Question

[Shit em không trôi]

Cho các số thực dương x,y,z thỏa mãn xyz = 1 CMR:

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{^{ }yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\ge3\sqrt{3}\)

Answer 1

[TFBoys]

Ta có \(x^3+y^3\ge xy\left(x+y\right)\)

\(\Rightarrow1+x^3+y^3\ge xyz+xy\left(x+y+z\right)=xy\left(x+y+z\right)\)

Tương tự ta có

\(VT\ge\dfrac{\sqrt{xy\left(x+y+z\right)}}{xy}+\dfrac{\sqrt{yz\left(x+y+z\right)}}{yz}+\dfrac{\sqrt{zx\left(x+y+z\right)}}{zx}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(=\sqrt{x+y+z}.\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\)

\(\ge\sqrt{3\sqrt[3]{xyz}}.\dfrac{3\sqrt[6]{xyz}}{1}=3\sqrt{3}\)

\("="\Leftrightarrow x=y=z=1\)