2. Let $\mathcal{F}$ be the set of continuous functions $f: \mathbb{R} \longrightarrow \mathbb{R}$ that satisfy the condition

$$
e^{f(x)}+f(x) \geq x+1
$$

for any real number $x$. Determine the minimum value that the integral

$$
I(f)=\int_{0}^{e} f(x) d x
$$

can take when $f$ ranges over $\mathcal{F}$.
## Solution:

We will show that the minimum value is $\frac{3}{2}$.

Consider the function $g: \mathbb{R} \longrightarrow \mathbb{R}, g(x)=e^{x}+x-1$.

This function is strictly increasing and continuous, with $\operatorname{Im}(g)=\mathbb{R}$, hence it is invertible,...........1p

with the inverse also being continuous and strictly increasing..................................... $\mathbf{p}$

The inequality in the statement can be written in the form $g(f(x)) \geq x, \forall x \in \mathbb{R}$, from which
$f(x) \geq g^{-1}(x), \forall x \in \mathbb{R}$

Since $g^{-1} \in \mathcal{F}$, and $I(f) \geq I\left(g^{-1}\right), \forall f \in \mathcal{F}$, the minimum value is $I\left(g^{-1}\right) \ldots$

With the substitution $t=g^{-1}(x)$, we have

$$
I\left(g^{-1}\right)=\int_{0}^{e} g^{-1}(x) d x=\int_{0}^{1} t g^{\prime}(t) d t=\left.\left((t-1) e^{t}+\frac{t^{2}}{2}\right)\right|_{0} ^{1}=\frac{3}{2}
$$

Note: The last calculation reenacts the proof of Young's theorem, which can also be invoked to obtain the conclusion.