let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$

Question

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?

Answer 1

Since for positive numbers $a,b$ we have $a+b\geq 2\sqrt{ab}$ we have also $$f(x) \geq 2$$ (put $a=x$ and $b=1/x$) with equality iff $x=1$.

Now $g(x)= (x+2)^2-10$, so for $x\geq -2$ it is increasing and thus the minimum of $g$ is at $x=2$.

So the minimum of $g(f(x))$ is at $f(x)=2$ i.e. $x=1$.

Answer 2

$f(x)$ is clearly an increasing function for $x\geq1$, and has a minimum value of 2 at $x=1$. Also g(x) is an increasing function for $x\geq-2$. Henceforth it is easy to see that $g(f(x))$ is an increasing function for the domain $x\geq1$ with the minimum value of 6 at $x=1$.

Answer 3

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 $

$(2+2)^2-10=6$.