\displaystyle \begin{align*} & 0<x<1\wedge f\left( x \right)=\int_x^1 \frac{\left( 1-t \right)^2}{t^2} \text{d}t \\ & \text{Prove}:\ \ f\left( x \right)\ge \frac{2\left( 1-x \right)^3}{3x\left( 1+x \right)} \end{align*}

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@CalvinLin How to accept answers? – Ryan Jan 3 '13 at 4:45
Go to the question that you posted. To the left of the answer that you want to accept, click on the check sign to the left, (and then the up vote if you wish). – Calvin Lin Jan 3 '13 at 4:47
Just click on the check mark icon underneath what you consider the best or most helpful answer. I'd also recommend "up voting" any answers you found helpful by clicking on the up arrows on the top left of an answer. – JohnD Jan 3 '13 at 4:47
@Ryan : Your way of using TeX is very strange. Writing {{{{a+b}}}} where a+b will suffice just makes things difficult for anyone trying to either edit it or understand the TeX. – Michael Hardy Jan 3 '13 at 4:58

Let $g(x) = \frac {2(1-x)^3}{3x(1+x)}$. We want to show that $f(x) \geq g(x)$. Since we have $f(1) = 0, g(1) = 0$, it suffices to show that $f'(x) \leq g'(x)$ for $x\in [0,1]$.
From definition of $f(x)$, we have $f'(x) = - \frac {(1-x)^2}{x^2}$. By Wolfram Alpha (or do this yourself), we have $g'(x) = -\frac {2 (-1+x)^2 (1+4 x+x^2)}{3 x^2 (1+x)^2}$. Hence, it suffices to show that $- \frac {(1-x)^2}{x^2} \leq -\frac {2 (-1+x)^2(1+4x + x^2}{3x^2(1+x)^2}$, or equivalently that $3(1+x)^2 \geq 2(1+4x+x^2)$ (Why is the inequality sign switched?), which reduces to $(1-x)^2 \geq 0$.
I'm sorry, I'm not seeing why we should have $f'(x)\leq g'(x)$. Could you explain this? – user50407 Jan 3 '13 at 11:00
@mr.FS It's not a necessary condition, but a sufficient (and somewhat strong) condition. If $f'(x) \leq g'(x)$ for $x\in [0,1]$ and $f(1)=g(1)$, then $f(x) \geq g(x)$ for $x\in [0,1]$. Simplest explanation will be to draw a picture. The inequality sign 'flips' because we're not moving to the right, and reaching the same end value. – Calvin Lin Jan 3 '13 at 16:18
@mr.FS It should have been "now moving to the right, and reaching the same end value". The more common version of this technique that you would have seen, is that if $f(0)=g(0)$ and $f' \geq g'$ then $f \geq g$. I'm using a variant of this. Drawing a picture really helps (even if it doesn't yet constitute a proper proof). – Calvin Lin Jan 3 '13 at 17:10