<s>[INST] let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$ 
Let $f(x)=x+\frac{1}{x}$ for all $x \geq 1$ and $g(x)=x^2+4x-6$. Find minimum of $g\circ f$.

My try:
The domain of $g\circ f$ is $[1,+\infty)$ and we have
$$g(f(x))=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)-6.$$
At $x=-2$, the minimum of $g(f(x))=-10$. Is that right?
 [/INST] 1)AM-GM:
$x>0$:
$x+1/x \ge 2 \sqrt{x(1/x)}=2$;
$f(1)=2=\min ${$f(x)| x \ge 1$}.
2) $g(x)=x^2+4x-6 =$
$(x+2)^2-10$ is increasing for $x \ge -2$.
3) $g(f(x))=(f(x)+2)^2-10 \ge $
$(2+2)^2-10=6$.
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