Answer
\[\int_{0}^{1}\sqrt{e^{2x}+1}\,dx\geq e-1\]
Work Step by Step
To show that:- \[\int_{0}^{1}\sqrt{1+e^{2x}}\;dx\geq e-1\]
Since on $[0,1]$ \[e^{2x}> 0\]
\[\Rightarrow e^{2x}+1> e^{2x}\]
\[\Rightarrow \sqrt{e^{2x}+1}> \sqrt{e^{2x}}\]
We know that if $f(x)\leq g(x)$ on $[a,b]$
Then \[\int_{a}^{b}f(x)\,dx\leq \int_{a}^{b}g(x)\,dx\]
\[\Rightarrow \int_{0}^{1}\sqrt{e^{2x}+1}\,dx>\int_{0}^{1}\sqrt{e^{2x}}\,dx=\int_{0}^{1}e^x \,dx\]
\[\Rightarrow \int_{0}^{1}\sqrt{e^{2x}+1}\,dx>\int_{0}^{1}e^x \,dx=\left[e^x\right]_{0}^{1}=e-1\]
Hence, \[\int_{0}^{1}\sqrt{e^{2x}+1}\,dx> e-1\]