Finding the maximum and minimum values of $f(x)=a^x+a^{1/x}$ Let $f(x)=a^x+a^{1/x}\ (x\gt 0)$ where $a\in\mathbb R$ is a constant.

Question 1 : What is the maximum value of $f(x)$ for $0\lt a\lt 1$?
Question 2 : What is the minimum value of $f(x)$ for $a\gt 1$?

The followings may be true:
Conjecture 1 : For $a=1/2$, the maximum value of $f(x)$ is $f(1)=1$.
Conjecture 2 : For $0\lt a\lt 1, a\not =1/2$, $f(x)$ does not have the maximum value.
Conjecture 3 : For $a\gt 1$, the minimum value of $f(x)$ is $f(1)=2a$.
I'm afraid that these may be very easy questions, but to be honest I don't have any good idea to solve them. Can anyone help?
Edit 1 : Question 2 has already been solved by Barry Cipra. (Conjecture 3 is true) However, Question 1 has not been solved yet.
Edit 2 : I've just been able to prove that the conjecture 1 is true. Also, I found that the conjecture 2 is not true, but I'm not able to answer the question 1 in general.
 A: Conjecture 3 can be handled with the arithmetic-geometric mean inequality, in the form $u+v\ge2\sqrt{uv}$:  If $a\gt1$, then
$$a^x+a^{1/x}\ge2\sqrt{a^xa^{1/x}}=2\sqrt{a^{x+1/x}}\ge2\sqrt{a^{2\sqrt{x\cdot1/x}}}=2\sqrt{a^2}=2a$$
with equality when $x=1$.  (Note, the assumption $a\gt1$ is crucial when replacing the exponent $x+1/x$ with the smaller exponent $2\sqrt{x\cdot1/x}$.  If $a$ were smaller than $1$, the inequality would point in the "wrong" direction.)
A: Whenever you're solving questions like these, it is always best to use calculus.
So we have a function where $$f(x) = a^x + a^{1/x}$$
Taking the derivative: $$f'(x)= a^x\ln a - \frac{a^{1/x}\ln a}{x^2}$$
Now we set the equation equal to 0 find the value(s) of x that make it maximum or minimum.
$$a^x\ln a - \frac{a^{1/x}\ln a}{x^2} = 0$$
$\ln a$ is always, greater than $0$, so 
$$a^x-\frac{a^{1/x}}{x^2} = 0$$
Therefore, when $x = -1, 1$, $f'(x) = 0$ EDIT: $x > 0$, so we can only use $x = 1$
When $a > 1$
Our minimum value occurs at $x = 1$, because from $(0, 1)$, $f'(x) < 0$, and from $(1, \infty)$, $f'(x) > 0$, using this graph (I choose an arbitrary a in order to graph it).
$$f(1)= a^{1} + a^{1} = 2a$$
When $a = 1/2$
We go back to our $f'(x)$ equation, and we substitute $a = 1/2$ We get $$\left(\frac{1}{2}\right)^{x} - \frac{\left(\frac{1}{2}\right)^{1/x}}{x^2}$$ Looking at this, we see that the max. value occurs at $x = 0.215106, 4.64886$, and the min. value occurs at $x = 1$.
When $0 < a < 1$, $a \not= 1/2$
You can graph the function for an arbitrary $a$, and you'll see that the function has a max value at $x = 1$
A: I've just been able to prove that the conjecture 1 is true (Conjecture 1 : For $a=1/2$, the maximum value of $f(x)=a^x+a^{1/x}$ is $f(1)=1$.
). Also, I found that the conjecture 2 is not true, but I'm not able to answer the question 1 in general.
In the following, I'm going to prove that the maximum value of $f(x)=a^x+a^{1/x}$ is $f(1)=1$ for $a=1/2$. 
Since setting $x=1/t$ in $a^x+a^{1/x}$ gives us $a^{1/t}+a^t$, we can see that it is sufficient to prove that the maximum value of $f(x)\ (0\lt x\le 1)$ is $1$. 
Since $f(x)\lt 1\iff (1-2^{-x})\cdot 2^{1/x}\gt 1$, I'm going to prove that $$g(x)=(1-2^{-x})\cdot 2^{1/x}\gt 1\ (0\lt x\lt 1).$$
We get $$g'(x)=\log2\cdot 2^{1/x}\cdot \frac{(x^2+1)\cdot 2^{-x}-1}{x^2}.$$
In the following, I'm going to prove that $h(x)=(x^2+1)\cdot 2^{-x}-1\lt 0$.
The tangent line of $y=2^x$ at $(0,1)$ is $y=(\log2)x+1$ and at $(1,2)$ is $y=(2\log2)(x-1)+2.$ Since the intersection point of these lines is $(2-1/\log2,2\log2)$, let 
$$h(x)=(\log2)x+1\ \ (0\lt x\lt 2-1/\log2), (2\log2)(x-1)+2\ \ (2-1/\log2\le x\lt 1).$$
Then, since $y=2^x$ is convex downward, we get $2^x\gt h(x).$ Also, since we can see $$(2-1/\log2)^2+1\lt 2\log 2\iff (\log4-1)(1-\log2)^2\gt 0,$$
we can get $h(2-1/\log 2)\gt x^2+1$. Hence, since $y=x^2+1$ is convex downward, we get $h(x)\gt x^2+1.$
Hence, since we get $2^x\gt x^2+1$, we get $(x^2+1)\cdot 2^{-x}-1\lt 0$. Hence, since we get $g'(x)\lt 0$ and $g(1)=1$, we get $g(x)\gt 1$ for $0\lt x\lt 1.$ Q.E.D.
