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: 
If we denote $s:=a+b+c$, we have to minimize
$$\frac{a}{s-a}+\frac{b}{s-b}+\frac{c}{s-c}$$
for $a,b,c>0$.

Using new variables $x:=a/s$, $y:=b/s$, $z:=c/s$ we can see, that this is equivalent to minimizing
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z}$$
for $x+y+z=1$.

If we notice that [the function](http://www.wolframalpha.com/input/?i=x%2F%281-x%29)
$$f(x)=\frac{x}{1-x}=-1+\frac1{1-x}=-1-\frac1{x-1}$$
is convex on the interval $(0,1)$, we have
$$\frac{f(x)+f(y)+f(z)}3 \ge f\left(\frac{x+y+z}3\right),$$
i.e.
$$f(x)+f(y)+f(z) \ge 3f(1/3).$$
By computing $f(1/3)=1/2$ we see, that the last inequality is precisely
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \ge \frac32.$$
\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 $x=y=z$.