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 $\frac{a+b+c}{c+a}$.
We can also use rearrangement inequality to prove it. 

$\displaystyle\frac a{b+c}+\frac b{a+c}+\frac c{a+b}$  

$\displaystyle=\frac12\left(\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)+\left(\frac a{b+c}+\frac b{a+c}+\frac c{a+b}\right)\right)$  

$\displaystyle\geq\frac12\left(\left(\frac a{a+c}+\frac b{a+b}+\frac c{b+c}\right)+\left(\frac a{a+b}+\frac b{b+c}+\frac c{a+c}\right)\right)$  

$=\dfrac32$