# prove inequality

Given that $s+t=p$, prove that $2s^2 \geq p^2 - 2t^2$.

Here's what I came up with:

\begin{align} 2s^2 \geq (s+t)^2 - 2t^2 \\ \implies 2s^2 \geq s^2 + 2st - t^2 \\ \implies s^2 - 2st + t^2 \geq 0 \\ \implies (s-t)^2 \geq 0 \end{align}

Is it enough to prove this inequality?

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Should change the $2st^2$ to $2t^2$ though (3th line). –  Bob Jan 30 '13 at 13:07

$$(s+t)^2 = p^2 \leq 2 \cdot (s^2+t^2) = 2 \cdot (s+t)^2 - 4st$$

Thus $$(s+t)^2-4st \geq 0 <=> (s-t)^2 \geq 0$$

Im not sure if you need an inverse proof here, this might suffice.

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