74th Romanian Mathematical Olympiad

Because $f^{\prime }(x)\ge 0$ for any $x\ge 0$, the function $f$ is monotonically increasing, so that $f(x)\ge f(0)=0$, for any $x\ge 0$.

Consider an arbitrary fixed $n\in {\mathbb{N}}^{\ast }$ and the function $F:[0,\infty )\to \mathbb{R}$ defined for any $x\ge 0$ by $$F(x)=(n+1)\cdot {\left({\int }_0^{x}f(t{)}^{n}dt\right)}^2-{\int }_0^{x}f(t{)}^{2n+1}dt.$$ We will show that $F$ is monotonically increasing, which, since $F(0)=0$, will prove the stated inequality.

The function $f$ being continuous, $F$ is differentiable and $$\begin{array}{rl}{F}^{\prime }(x)& =2(n+1)\cdot \left({\int }_0^{x}f(t{)}^{n}dt\right)\cdot f(x{)}^n-f(x{)}^{2n+1}\\ & =f(x{)}^n\cdot \left(2(n+1)\cdot \left({\int }_0^{x}f(t{)}^{n}dt\right)-f(x{)}^{n+1}\right).\end{array}$$ Because $f^{\prime }(x)\le 1$, for any $x\ge 0$, we have that $f(x{)}^n\ge f(x{)}^n\cdot f^{\prime }(x)$, $\forall x\ge 0$, hence $$(n+1)\cdot \left({\int }_0^{x}f(t{)}^{n}dt\right)\ge {\int }_0^x(n+1)\cdot f(t{)}^n\cdot f^{\prime }(t)dt=f(x{)}^{n+1}.$$ It follows that $$\begin{array}{rl}{F}^{\prime }(x)& =f(x{)}^n\cdot \left(2(n+1)\cdot \left({\int }_0^{x}f(t{)}^{n}dt\right)-f(x{)}^{n+1}\right)\ge \\ & \ge f(x{)}^n\cdot \left(2\cdot f(x{)}^{n+1}-f(x{)}^{n+1}\right)=f(x{)}^{2n+1}\ge 0,\end{array}$$ for any $x\ge 0$. Thus, $F$ is increasing, and so $F(x)\ge F(0)=0$, for any $x\ge 0$. We obtain the stated inequality.