Prove integral inequality: $\int_{0}^{\frac{\pi}{2}}e^{\sin x}\,dx\geq\frac{\pi}{2}(e-1)$

Note that $\sin x \geq \frac 2 \pi x$ holds for $x \in [0, \frac \pi 2]$. Thus $$\int_{0}^{\pi/2}e^{\sin x}dx \geq \int_{0}^{\pi/2}e^{\frac 2 \pi x}dx =\frac \pi 2 (e-1)$$

Since @sera already provided a good and simple answer, this is a loog comment.

Your idea of using Taylor series was good but it should have been $$e^{\sin(x)}=\sum_{n=0}^\infty \frac {\sin^n(x)} {n!}\implies \int_0^{\frac \pi 2} e^{\sin(x)}\, dx=\sum_{n=0}^\infty \frac {1} {n!}\int_0^{\frac \pi 2}\sin^n(x)\,dx$$ Since $$\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt{\pi }}2 \frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma\left(\frac{n}{2}+1\right)}$$ So, consider $$S_p=\frac{\sqrt{\pi }}2\sum_{n=0}^p \frac{ \Gamma \left(\frac{n+1}{2}\right)}{n! _, \Gamma\left(\frac{n}{2}+1\right)}$$ and remember that $\frac \pi 2 (e-1) <e $

the partial sums (which are increasing) generate the sequence $$\left\{\frac{\pi }{2},1+\frac{\pi }{2},1+\frac{5 \pi }{8}, \frac{10}{9}+\frac{5 \pi}{8},\frac{10}{9}+\frac{81 \pi }{128}, \frac{251}{225}+\frac{81 \pi}{128},\frac{251}{225}+\frac{2917 \pi }{4608}\right\} $$ and the third term is already larger than the rhs