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

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 $(1-x)^2 \geq 0$.
