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

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

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Equality is reached when $f(t)=t$. It is natural to set $f(t)=t+g(t)$. Then, everything can be stated in terms of positivity conditions : $\int_{x}^{1} g(t)dt \geq 0$ for every $x$, and we must show $\int_{0}^{1} g(t)(g(t)+2t)dt \geq 0$. –  Ewan Delanoy Oct 26 '12 at 4:51

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: $$\int_0^1 f(t)^2dt\cdot\int_0^1 t^2dt\ge\left[\int_0^1tf(t)dt\right]^2.$$