International Mathematics Competition
for University Students
2019

IMC2019: Day 1, Problem 3

Problem 3. Let \(\displaystyle f:(-1,1)\to\RR\) be a twice differentiable function such that

\(\displaystyle {2f'(x)+xf''(x)\geq1} \quad\text{for \(\displaystyle x\in(-1,1)\)}. \)

Prove that

\(\displaystyle \int_{-1}^1xf(x)\,\mathrm{d}x\geq\frac{1}{3}. \)

Proposed by Orif Ibrogimov, ETH Zurich and National University of Uzbekistan and

Karim Rakhimov, Scuola Normale Superiore and National University of Uzbekistan

Solution. Let

\(\displaystyle g(x) = xf(x) -\frac{x^2}{2}. \)

Notice that

\(\displaystyle g''(x) = 2f'(x)+xf''(x)-1 \ge0, \)

so \(\displaystyle g\) is convex. Estimate \(\displaystyle g\) by its tangent at \(\displaystyle 0\): let \(\displaystyle g'(0)=a\), then

\(\displaystyle g(x) = g(0)+g'(0)x = ax \)

and therefore

\(\displaystyle \int_{-1}^1xf(x)\,\mathrm{d}x = \int_{-1}^1\Bigl(g(x)+\frac{x^2}{2}\Bigr)\,\mathrm{d}x \geq \int_{-1}^1 \Bigl( ax + \frac{x^2}{2} \Bigr)\,\mathrm{d}x = \frac{1}{3}. \)