Question

Cho x,y,z>0 thoã mãn: x³+y³+z³=1

Chứng minh rằng: \(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)

Answer 1

Ta có:

\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)

Áp dụng BĐT Cosi ta có:

\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)

Cmtt:

\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)

\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)

\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)