Ta có \(x\leq y-1\Rightarrow 0<\frac{x}{y^{2}}\leq \frac{1}{y}-\frac{1}{y^{2}}=\frac{1}{4}-(\frac{1}{y}-\frac{1}{2})^{2}\leq \frac{1}{4}\)
Đặt \(t=\frac{x}{y^{2}}\Rightarrow 0
Ta có: \(T=\frac{\frac{x}{y^{2}}+3}{\sqrt{(\frac{x}{y^{2}})^{2}+1}}-\frac{1}{5}.\frac{2.\frac{x}{y^{2}}+1}{\frac{x}{y^{2}}+1}\Rightarrow T=f(t)=\frac{t+3}{\sqrt{t^{2}+1}}-\frac{1}{5}.\frac{2t+1}{t+1}\) với \(0