\(\frac{t^2}{2t^2+3}+\frac{2}{1+t}-\frac{34}{33}=\frac{-35t^3+97t^2-102t+96}{33\left(t+1\right)\left(2t^2+3\right)}=\frac{\left(2-t\right)\left(35t^2-27t+48\right)}{33\left(t+1\right)\left(2t^2+3\right)}\ge0\) \(\forall t\in\left[1;2\right]\)
\(\Rightarrow\frac{t^2}{2t^2+3}+\frac{2}{1+t}\ge\frac{34}{33}\)
Dấu "=" xảy ra khi \(t=2\)