Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$ Show that: $ \frac{x^2}{2(x + 1)} < x − \ln(x + 1) < \frac{x^2}{ 2} , x > 0$
One method is considering a function $f(x) = x − \ln(1 + x) − \frac{x^2}{2}$ and showing its derivative $f'(x) < 0$.
Similarly, considering another function $g(x) = x − \ln(1 + x) − \frac{x^2}{ 2(x + 1)}$ and showing its $g'(x) > 0$. 
Is there some other method that is not lengthy?

You can use limits for the natural log (which can basically be seen after Taylor expansion, proof is  here) to see that 
$\ln(x+1) > x - x^2/2$ and $(x+1)\ln(x+1) < x + x^2/2$ 
The first inequality  immediately gives  
$x-\ln(x+1) < x^2/2$  and 
and the second one gives 
$(x+1) (x-\ln(x+1)) > x + x^2 - (x + x^2/2) = x^2/2$ 
so we have the two required limits.
