Đặt \(\left(\frac{1}{x};\frac{1}{y}\right)=\left(a;b\right)\Rightarrow ab+a+b=3\)
\(\Rightarrow ab+2\sqrt{ab}\le3\Rightarrow\left(\sqrt{ab}+3\right)\left(\sqrt{ab}-1\right)\le0\)
\(\Rightarrow\sqrt{ab}\le1\Rightarrow ab\le1\)
\(P=\frac{a}{\sqrt{3+a^2}}+\frac{b}{\sqrt{3+b^2}}=\frac{a}{\sqrt{ab+a+b+a^2}}+\frac{b}{\sqrt{ab+a+b+b^2}}\)
\(=\frac{a}{\sqrt{\left(a+b\right)\left(a+1\right)}}+\frac{b}{\sqrt{\left(a+b\right)\left(b+1\right)}}\le\frac{1}{2}\left(\frac{a}{a+b}+\frac{a}{a+1}+\frac{b}{a+b}+\frac{b}{b+1}\right)\)
\(P\le\frac{1}{2}\left(1+\frac{a}{a+1}+\frac{b}{b+1}\right)=\frac{1}{2}\left(1+\frac{ab+a+ab+b}{ab+a+b+1}\right)=\frac{1}{2}\left(1+\frac{ab+3}{4}\right)\)
\(P\le\frac{1}{2}\left(1+\frac{1+3}{4}\right)=1\)
Dấu "=" xảy ra khi \(a=b=1\) hay \(x=y=1\)
