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
$$\frac{x}{1-x}+\frac{y}{1-y}+\frac{z}{1-z} \ge \frac32.$$
Here is another nice proof.

We have to prove;

$$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}\\\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}\geq \frac{3}{2}+3\\(a+b+c)\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq \frac{9}{2}\\([b+c]+[a+c]+[a+b])\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\geq 9\\$$Now we use this well known inequality for $x>0\to x+\frac{1}{x}\geq2$;

(quick proof: $\left (\sqrt x-\frac{1}{\sqrt x}\right )^2\geq0\to x+\frac1x\geq2$)
$$1+1+1+\left[\frac{a+b}{a+c}+\frac{a+c}{a+b}\right]+\left[\frac{a+b}{b+c}+\frac{b+c}{a+b}\right]+\left[\frac{a+c}{b+c}+\frac{b+c}{a+c}\right]\geq 9\\q.e.d.$$