2019 IMC #3

Problem: Let $f:(-1,1)\to \mathbb{R}$ be a twice differentiable function such that $$2f’(x)+xf''(x)\geqslant 1 \quad \text{ for } x\in (-1,1).$$Prove that $$\int_{-1}^{1}xf(x)dx\geqslant \frac{1}{3}.$$

Solution 1: The LHS of the original inequality looks like the result of an application of the product rule. Indeed, the derivative of $f'g$ is $g'f'+gf''$. We cannot make $g=x$ and $g'=2$, but we can multiply by $x$ to get $g=x^2, g'=2x$. Now to do this we need to be careful about the sign of $x$. So, suppose that $x > 0$. Then $2xf'(x)+x^2f''(x) \ge x$ and hence $(x^2f')' \ge x$ so that $x^2f'(x) \ge \frac{1}{2} x^2$ and hence $f'(x) \ge \frac{1}{2}$. Similarly, we know that $f'(x)\ge \frac{1}{2}$ for $x < 0$, so we have proved this for all $x$ (for $x=0$ by continuity or by just plugging in $x=0$ into the original inequality). The rest is easy. WLOG, assume that $f(0)=0$ since shifting $f$ by a constant changes neither the integral nor the condition on the derivatives (here we make use of $\int_{-1}^1 xdx=0$). Then $f(x) \ge \frac{x}{2}$ for $x\ge 0$ and $f(x)\le \frac{x}{2}$ for $x<0$ so that $xf(x) \ge \frac{x^2}{2}$ for all $x$ and we have \[\int_{-1}^1 xf(x) dx \ge \int_{-1}^1 \frac{x^2}{2} dx=\frac{1}{3},\] as desired. $\square$

Solution 2: Let $g(x)=xf(x)-\frac{x^2}{2}$. Then, we have $g''(x) \ge 0$. So, note that $$\int_{-1}^{1}g(x)dx\geq 2\cdot g\left(\frac{1+(-1)}{2}\right)=0$$

$$\implies \int_{-1}^{1}xf(x)dx\geqslant \int_{-1}^{1}\frac{x^2}{2}=\frac{1}{3},$$ as desired. $\square$