Đặt \(\frac{1}{b}=c\Rightarrow1\ge a+c\ge2\sqrt{ac}\Rightarrow ac\le\frac{1}{4}\)
\(T=\frac{a.\frac{1}{c}}{a^2+\frac{1}{c^2}}=\frac{ac}{a^2c^2+1}=\frac{1}{ac+\frac{1}{ac}}=\frac{1}{ac+\frac{1}{16ac}+\frac{15}{16ac}}\le\frac{1}{2\sqrt{ac.\frac{1}{16}ac}+\frac{15}{16.\frac{1}{4}}}=\frac{4}{17}\)
Dấu "=" xảy ra khi \(a=c=\frac{1}{2}\) hay \(\left\{{}\begin{matrix}a=\frac{1}{2}\\b=2\end{matrix}\right.\)