# Showing a function is negative

I have the following function $f:[0,\frac{1}{2}] \to \mathbb{R}$:

$$f(p) = p^2(\log(p))^2 - (1-p)^2(\log(1-p))^2 + (1-2p)\log(p)\log(1-p) + (1-2p)\{p\log(p)+(1-p)\log(1-p)\}$$

The inequality I need to show is $$f(p) \leq 0$$I can show that $f(0) = f(1/2) = 0$, and that $f'(0) = -1$, $f'(1/2) = 0$. The graph of $f$ looks like

.

Since its not monotonic/convex/concave I'm stuck. Any leads are welcome!

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The domain of $f$ doesn't include $0$ because of the $\log(p)$ terms. How did you get $f(0)=0$? – Patrick Apr 11 '12 at 0:16
In the limit as p goes to 0.. I should have been more clear though. – VSJ Apr 11 '12 at 0:57
Can you prove that $f'$ has only one zero? – lhf Apr 11 '12 at 2:30
@lhf: I tried that, but the derivative did not look manageable enough. – VSJ Apr 11 '12 at 2:46
Using laws of logs, you can simplify the equation. You might be then able to take the derivative, as @lhf suggested. – Ben Apr 19 '12 at 8:55