Integral inequality with exponents

Question

Let $f(x):[0;1]\to\mathbb{R}$ be continuous function. Prove that $$\int_0^1e^{f(x)}dx \cdot \int_0^1e^{-f(x)}dx \geq 1+\int_0^1(f(x))^2dx-\Bigg(\int_0^1f(x)dx\Bigg)^2.$$
I tried to use Cauchy–Schwarz inequality. I tried some substitutions and integration by parts. I also tried to see on the following inequality $\int_0^1\big(e^{f(x)}-g(x)\big)^2dx\geq0$ and find such a function $g(x)$ to get closer to initial inequality. I don't know what else I should try.

Answer 1

Now I know the solution. Lemma: $\forall t \in \mathbb{R}$ the following inequality holds $e^t+e^{-t} \geq 2+t^2$. Proof of the Lemma: $$e^t+e^{-t} = 2 \sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!}=2(1+\frac{t^2}{2}+...) \geq 2+t^2$$ Now put $t=f(x)-f(y)$. We have $$e^{f(x)-f(y)}+e^{f(y)-f(x)} \geq 2+(f(x))^2+(f(y))^2-2f(x)f(y)$$ Taking double integral with respect to $x$ and $y$ over $[0 ; 1]^2$ we have $$\int_{0}^{1}\int_{0}^{1}e^{f(x)}e^{-f(y)}dxdy+\int_{0}^{1}\int_{0}^{1}e^{f(y)}e^{-f(x)}dxdy\geq2+\int_{0}^{1}(f(x))^2dx+\int_{0}^{1}(f(y))^2dy-\int_{0}^{1}\int_{0}^{1}2f(x)f(y)dxdy$$ We can obviously represent double integral as a product of two integrals in this case. After dividing by $2$ we get the desired inequality.