Đặt \(2^{x^2}=t\ge1\)
\(\Rightarrow t^2-2\left(m+1\right)t+m+3\ge0\)
\(\Leftrightarrow t^2-2t+3\ge\left(2t-1\right)m\)
Do \(t\ge1\Rightarrow2t-1>0\) nên BPT tương đương:
\(m\le\dfrac{t^2-2t+3}{2t-1}\)
\(\Leftrightarrow m\le\min\limits_{t\ge1}f\left(t\right)\) với \(f\left(t\right)=\dfrac{t^2-2t+3}{2t-1}\)
\(f'\left(t\right)=\dfrac{\left(2t-2\right)\left(2t-1\right)-2\left(t^2-2t+3\right)}{\left(2t-1\right)^2}=\dfrac{2t^2-2t-4}{\left(2t-1\right)^2}=0\)
\(\Rightarrow t=2\)
\(f\left(1\right)=2\) ; \(f\left(2\right)=1\Rightarrow\min\limits_{t\ge1}f\left(t\right)=1\)
\(\Rightarrow m\le1\)