Question

cho a,b dương tỏa mãn: \(\sqrt{a}+\sqrt{b}\ge2\)

CMR: \(\sqrt{a^3}+\sqrt{b^3}\ge a+b\)

Answer 1

\(\left\{{}\begin{matrix}\sqrt{a}=x\ge0\\\sqrt{b}=y\ge0\end{matrix}\right.\Rightarrow x+y\ge2\)

Ta cm bđt sau: \(2\left(x^3+y^3\right)\ge\left(x+y\right)\left(x^2+y^2\right)\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\ge0\) (đúng)

Mà: \(x+y\ge2\Rightarrow x^3+y^3\ge x^2+y^2\)

\(\Rightarrow\sqrt{a^3}+\sqrt{b^3}\ge a+b\)(đpcm)

\("="\Leftrightarrow a=b=1\)