If $\int_x^1f(t)dt\ge\frac{1-x^2}2$, $x\in[0,1]$, prove that $\int_0^1f(t)^2dt\ge1/3$.

Question

Let $f$ be continuous on $[0,1]$ that satisfies

$\int_x^1f(t)dt\ge\frac{1-x^2}2,x\in[0,1]$.

Prove that $\int_0^1f(t)^2dt\ge1/3$.

Answer 1

Define $F(x)=\int_x^1f(t)dt$. Integration by parts implies that $$\int_0^1tf(t)dt=\int_0^1F(t)dt\ge\dfrac{1}{3}.$$ Then the conclusion follows from Cauchy's inequality: $$\int_0^1 f(t)^2dt\cdot\int_0^1 t^2dt\ge\left[\int_0^1tf(t)dt\right]^2. $$