Question: How to find the limit 

$$\lim\_{x \rightarrow \infty} x^{2/3} -(x^2+1)^{1/3}$$

I've tried: 

$$\lim x^{2/3} -(x^2+1)^{1/3} = \lim x^{2/3} \cdot \lim \left[1 - \left(1+\frac{1}{x^2}\right)^{1/3} \right] = \infty \cdot 0$$

Any help, please?

Answer: 
Use AM > HM on $\frac{a+b+c}{a+b}, \frac{a+b+c}{b+c}$ and $\frac{a+b+c}{c+a}$.
\begin{eqnarray\*}
 % \nonumber to remove numbering (before each equation)
 && \frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}-\frac{3}{2}\\
 &=& \frac{2(x^3+y^3+z^3)-(x^2y+xy^2+y^2z+yz^2+z^2x+zx^2)}{2 (x + y) (x + z) (y + z)}\\
 &=& \frac{(x+y)(x-y)^2+(y+z)(y-z)^2+(z+x)(z-x)^2}{2 (x + y) (x + z) (y + z)}\geq0
 \end{eqnarray\*}
with equality holds if and only if $x=y=z$.