# Prove the following Integral Inequality.

Problem: Let $f(t)$ be a non-negative function defined on the interval $[0,1]$, and let $$\frac{f(t_1)+f(t_2)}{2}\leq f\left(\frac{t_1+t_2}{2}\right)$$ where $t_1,t_2\in[0,1].$ Prove the following inequalities: $$\int_{0}^{1}f(t)dt\leq3^n\int_{0}^{1}t^nf(t)dt, n\in\{0,1,2,3...\}$$ $$\frac{2}{(n+1)(n+2)}\int_{0}^{1}f(t)dt\leq\int_{0}^{1}t^nf(t)dt\leq\frac{2}{n+2}\int_{0}^{1}f(t)dt$$

My attempt: I tried to approach the first part through induction: Let $P(n)$ be the proposition that $$\int_{0}^{1}f(t)dt\leq3^n\int_{0}^{1}t^nf(t)dt.$$ Then $P(0)$ is trivially true. We'll try to prove $P(1),$ which implies $$\int_{0}^{1}f(t)dt\leq\int_{0}^{1}(3t)f(t)dt\Rightarrow\int_{0}^{1}(3t-1)f(t)dt\geq0.$$ Why is this true? I don't know. However, if we assume it to be true alongwith $P(k)$ we get $$\int_{0}^{1}f(t)dt\leq\int_{0}^{1}(3t)^kf(t)dt\Rightarrow\int_{0}^{1}((3t)^k-1)f(t)dt\geq0.$$ On multiplying by $3t$ on both sides of the inequality we get: $$\int_{0}^{1}((3t)^{k+1}-3t)f(t)dt\geq0.$$ Or, $$\int_{0}^{1}(3t)f(t)dt\leq\int_{0}^{1}(3t)^{k+1}f(t)dt.$$ But from $P(1)$ we can deduce that $$\int_{0}^{1}f(t)dt\leq\int_{0}^{1}(3t)f(t)dt$$ and thus, $P(k+1)$ holds, which completes the proof.

I tried using induction on the second inequality but failed. So I guess, I need help in proving $P(1)$ and the second inequality. Thanks.

• Note that from $\int_0^1 g(t)\ge 0$ it doesn't follow that $\int_0^1 tg(t)\ge 0.$ Moreover, you haven't used the hypothesis on $f$ anywhere. – mfl Oct 31 '16 at 10:05