Putnam 1991 
Topic: Calculus Analysis

Problem: Find the maximum value of \[ \int_0^y \sqrt{x^4 + (y-y^2)^2}\,dx \] for $0 \leq y \leq 1$.

Solution: For $0 \leq y \leq 1$ let $I(y) = \int_0^y \sqrt{x^4 + (y-y^2)^2}\,dx$. Claim: $I'(y) \geq 0$ with equality only in the (clearly non-optimal) case $y = 0$.

To see this, observe that

\[ I'(y) = \sqrt{y^4 + (y-y^2)^2} + \int_0^y \frac{(y-y^2)(1-2y)}{\sqrt{x^4 + (y-y^2)^2}}\,dx. \]

If $0 < y \leq 1/2$ clearly $I'(y)$ is positive. So suppose $y > 1/2$. Then $I'(y) > 0$ is equivalent to

\[ \sqrt{y^4 + (y-y^2)^2} > (y-y^2)(2y-1)\int_0^y \frac{dx}{\sqrt{x^4 + (y-y^2)^2}}. \]

Since

\[ \int_0^y \frac{dx}{\sqrt{x^4 + (y-y^2)^2}} \leq \int_0^y \frac{dx}{\sqrt{(y-y^2)^2}} = \frac{y}{y-y^2}, \]

it suffices to show

\[ \sqrt{y^4 + (y-y^2)^2} \geq (2y-1)y, \quad 1/2 \leq y \leq 1. \] This is the same as

\[ y^2 + (1-y)^2 \geq (2y-1)^2 \]

\[ \iff 2y^2 - 2y + 1 \geq 4y^2 - 4y + 1 \]

\[ \iff 2y \geq 2y^2, \]

 the last of which is clearly true.

Now, for $y < 1$, $I(y) < I(1) = \int_0^1 x^2\,dx = 1/3$, so $\boxed{1/3}$ is the maximum.