Let $f(x):[0, 1] \rightarrow R$ be an $m$-strongly convex function and $\mu$ be probability measure on $[0,1].$ For any $t<1$, the gaol is to find lowerbound on $\int_{0}^t f^2(x) d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound $$\int_{0}^t f^2(x) dx \ge \frac{ m^2 t^4}{36} \mu[0,t].$$ We do not know if our bound is tight. Moreover, our proof is really long and messy. A clean/simple proof of such an elementary result would be helpful.