Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Should change the $2st^2$ to $2t^2$ though (3th line). – Bob Jan 30 '13 at 13:07
up vote 1 down vote accepted

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

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.

share|cite|improve this answer

I think you have already got the answer. To write it up you just need to reverse your progress.

share|cite|improve this answer

Yes it is enough to prove it,if you go in the reverse direction you will get to know why it is true.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.