Problem 1. Let $a$ be a strictly positive real number. Determine the minimum value of the expression

$$
\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}-(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x
$$

as $f$ ranges over the set of concave functions $f:[0,1] \rightarrow \mathbb{R}$, with $f(0)=1$.
Solution. Let $f:[0,1] \rightarrow \mathbb{R}$ be a concave function such that $f(0)=1$. Then

$$
x^{a} f(x)+1-x^{a}=x^{a} f(x)+\left(1-x^{a}\right) f(0) \leq f\left(x^{a} \cdot x+\left(1-x^{a}\right) \cdot 0\right)=f\left(x^{a+1}\right)
$$

by multiplying with $(a+1) x^{a}$, we get

$$
(a+1) x^{2 a} f(x)+(a+1)\left(x^{a}-x^{2 a}\right) \leq(a+1) x^{a} f\left(x^{a+1}\right)
$$

for any $x$ in $[0,1]$.

Integrating over the interval $[0,1]$, we obtain

$$
(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x+(a+1) \int_{0}^{1}\left(x^{a}-x^{2 a}\right) \mathrm{d} x \leq(a+1) \int_{0}^{1} x^{a} f\left(x^{a+1}\right) \mathrm{d} x
$$

thus

$$
(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x+\frac{a}{2 a+1} \leq \int_{0}^{1} f(x) \mathrm{d} x
$$

$$
\begin{aligned}
& \text{Since} \int_{0}^{1} f(x) \mathrm{d} x \leq\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}+\frac{1}{4}, \text{ we get } \\
& \quad\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}-(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x \geq=\frac{a}{2 a+1}-\frac{1}{4}=\frac{2 a-1}{8 a+4}
\end{aligned}
$$

Since the function $f:[0,1] \rightarrow \mathbb{R}, f(x)=1-x$, is concave, $f(0)=1$ and

$$
\left(\int_{0}^{1} f(x) \mathrm{d} x\right)^{2}-(a+1) \int_{0}^{1} x^{2 a} f(x) \mathrm{d} x=\frac{2 a-1}{8 a+4}
$$

it follows that the required minimum is $(2 a-1) /(8 a+4)$.