Question

Cho \(a,b>0\) và \(a^2b+ab^2+ab=a^2+b^2\). Tìm max P \(=\) \(\frac{1}{a}\sqrt{1+\frac{a}{b}}+\frac{1}{b}\sqrt{1+\frac{b}{a}}\)

Answer 1

\(\frac{1}{2}\left(a+b\right)^2\le a^2+b^2=ab\left(a+b\right)+ab\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\frac{1}{2}\left(a+b\right)^2\le ab\left(a+b\right)+\frac{1}{4}\left(a+b\right)^2\)

\(\Rightarrow\frac{1}{4}\left(a+b\right)^2\le ab\left(a+b\right)\Rightarrow a+b\le4ab\)

\(\Rightarrow\frac{a+b}{ab}\le4\)

\(P=\frac{\sqrt{b\left(a+b\right)}}{ab}+\frac{\sqrt{a\left(a+b\right)}}{ab}=\frac{1}{2\sqrt{2}}\left(\frac{2\sqrt{2b\left(a+b\right)}+2\sqrt{2a\left(a+b\right)}}{ab}\right)\)

\(P\le\frac{1}{2\sqrt{2}}\left(\frac{2b+a+b+2a+a+b}{ab}\right)=\sqrt{2}\left(\frac{a+b}{ab}\right)\le4\sqrt{2}\)

\(P_{max}=4\sqrt{2}\) khi \(a=b=\frac{1}{2}\)

Answer 2

@Nguyễn Lê Phước Thịnh