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: 
By Cauchy-Schwarz $$\begin{align}\sum\_{cyc}\frac{a}{b+c}&=\sum\_{cyc}\left(\frac{a}{b+c}+1\right)-3\\
&=\sum\_{cyc}a\sum\_{cyc}\frac{1}{b+c}-3\\
&=\frac{1}{2}\sum\_{cyc}(b+c)\sum\_{cyc}\frac{1}{b+c}-3\\
&\geq\frac{1}{2}\cdot9-3=\frac{3}{2}.\end{align}$$
Done!

Another way.

By C-S $$\sum\limits\_{cyc}\frac{a}{b+c}=\sum\limits\_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c)^2}{\sum\limits\_{cyc}(ab+ac)}=\frac{(a+b+c)^2}{2(ab+ac+bc)}\geq\frac{3}{2},$$
where the last inequality it's just $\sum\limits\_{cyc}(a-b)^2\geq0$.
Done!

Another way.
$$\sum\limits\_{cyc}\frac{a}{b+c}-\frac{3}{2}=\sum\limits\_{cyc}\left(\frac{a}{b+c}-\frac{1}{2}\right)=\sum\limits\_{cyc}\frac{2a-b-c}{2(b+c)}=\sum\limits\_{cyc}\frac{a-b-(c-a)}{2(b+c)}=$$
$$=\sum\limits\_{cyc}(a-b)\left(\frac{1}{2(b+c)}-\frac{1}{2(a+c)}\right)=\sum\limits\_{cyc}\frac{(a-b)^2}{2(a+c)(b+c)}\geq0.$$
Done!
\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 $x=y=z$.