\(a+\frac{1}{b}\le1=>ab+1\le b\)
\(b\le ab+1\ge2\sqrt{ab}=>\sqrt{b}\ge2\sqrt{a}=>\frac{b}{a}\ge4\)
\(T=\frac{ab}{a^2+b^2}=\frac{1}{\frac{a}{b}+\frac{b}{a}}=\frac{1}{\frac{a}{b}+\frac{b}{16a}+\frac{15b}{16a}}\)
áp dụng cô si
\(\frac{a}{b}+\frac{b}{16a}\ge2\sqrt{\frac{ab}{16ab}}=\frac{1}{2}=>T\le\frac{1}{\frac{1}{2}+\frac{15}{16}.4}=\frac{4}{17}\)
\(=>MaxT=\frac{4}{17}\)
dấu = xảy ra khi
\(b=4a;\frac{a}{b}=\frac{b}{16a};ab=1\)
\(=>\hept{\begin{cases}4a^2=1\\b=4a\end{cases}=>\hept{\begin{cases}a=\frac{1}{2}\\b=2\end{cases}}}\)
