MathDB
Cute integral inequality

Source: Romanian District Olympiad 2024 12.4

March 10, 2024
Integralinequalitiesreal analysis

Problem Statement

Let f:[0,)Rf:[0,\infty)\to\mathbb{R} be a differentiable function, with a continous derivative. Given that f(0)=0f(0)=0 and 0f(x)10\leqslant f'(x)\leqslant 1 for every x>0x>0 prove that1n+10af(t)2n+1dt(0af(t)ndt)2,\frac{1}{n+1}\int_0^af(t)^{2n+1}\mathrm{d}t\leqslant\left(\int_0^af(t)^n\mathrm{d}t\right)^2,for any positive integer nn{} and real number a>0.a>0.