2019 ROMANIAN MATHEMATICAL OLYMPIAD

The required minimum is $(2a-1)/(8a+4)$ and is achieved for $f:[0,1]\to \mathbb{R}$, $f(x)=1-x$. The verification is routine and hence omitted.

Fix a concave function $f:[0,1]\to \mathbb{R}$ such that $f(0)=1$ to write $$x^{a}f(x)+1-x^a=x^{a}f(x)+(1-x^a)f(0)\le f(x^a\cdot x+(1-x^a)\cdot 0)=f(x^{a+1}),$$ for all $x$ in $[0,1]$. Multiply both sides by $(a+1)x^a$ and integrate over $[0,1]$ to get $$\begin{gathered} (a+1){\int }_0^1{x}^{2a}f(x)dx+\frac{a}{2a+1}\le {\int }_0^1(a+1)x^{a}f(x^{a+1})dx={\int }_0^{1}f(x)dx \\ \le {\left({\int }_0^{1}f(x)dx\right)}^2+\frac{1}{4}, \end{gathered}$$ and conclude that $${\left({\int }_0^{1}f(x)dx\right)}^2-(a+1){\int }_0^1{x}^{2a}f(x)dx\ge \frac{a}{2a+1}-\frac{1}{4}=\frac{2a-1}{8a+4}.$$