Find infimum of the sets of number $x + \frac{1}{x} $ Let $A = \{ z = x + \frac{1}{x} : x > 0 \} $ and $B = \{z = 2^x + 2^{1/x} : x > 0 \} $
I want to find $\inf A $ and $\inf B $.
Proof.
Clearly, by AM-GM inequality one has $x + \dfrac{1}{x} \geq 2 $ and $2^x + 2^{1/x} \geq 2 \sqrt{2^{x+1/x} } \geq 2 \sqrt{2^2} = 4 $.
Thus: claim $\inf A = 2 $ and $\inf B = 4 $
For the first one, we need to see that if $l$ is lower bound for $A$:
$$ x + \dfrac{1}{x} \geq l $$
for all $x$, then $2 \geq l$. Well, trivial: put $x=1$ then $1+1 \geq l $. So $\boxed{\inf A = 2 }$. Similarly, if $x=1$ in $B$ we see $2+2 \geq l$ so $\boxed{\inf B = 4}$ . QED
Is this enough work for the proof?

Looks good to me. In fact, you  prove not only that $\inf A=2$, but also that $\min A=2$. Similarly, you have proved that $\min B=4$. 
