Prove $a^x+a^{1/x}\le 2a$ ($\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$ 
Prove $a^x+a^{1/x}\le 2a$ (with $\frac{1}{2}\le a < 1$) for any $x\in (0,\infty)$.

Is it true? I can't figure out how to prove it.
It may use the conclusion $\ln x+\frac{1-x}{x}\ge 0, \forall x\in (0,+\infty) $ as this question is followed.

UPDATED:
I have find a complicated method to prove it.
Let $f(x) = a^x+a^{1/x}-2a$, with $\frac{1}{2}\le a<1, x\in(0,\infty)$.
$\therefore f'(x)=a^{1/x}\ln \frac{1}{a}(\frac{1}{x^2}-a^{x-1/x})$
we have $a^{1/x}\ln\frac{1}{a}>0$, so we can only consider
$g(x)=\frac{1}{x^2}-a^{x-1/x}, x>0, \frac{1}{2}\le a<1$
we can make the observation that $g(1)=0$, but we are not sure if it has any other roots. So we have to do some careful analysis.
If we keep taking derivative, we will get into endless trouble.
Note that $g(x)=\frac{1}{x^2}-a^{x-1/x}=a^{-2\log_a x}-a^{x-1/x}$
because $a<1$, so $a^x$ is decreasing, we can consider $h(x)=-2\log_a x-(x-\frac{1}{x})=-2\log_a x-x+\frac{1}{x},x>0,\frac{1}{2}\le a<1$
This is much easier. $h'(x)=-\dfrac{x^2+\dfrac{2}{\ln a}\cdot x+1}{x^2}$
because $\frac{1}{2}\le a<1$, so $\frac{2}{\ln a}\le -\frac{2}{\ln 2}\approx -2.9$. For the quadratic functions in the numerator, $\Delta >0$, there are two positive zeros $x_1,x_2$ and $0<x_1<1<x_2$, we have:
In $(0,x_1)$ and $(x_2,\infty)$，$h'(x)<0$, so $h(x)$ is decreasing.
In $(x_1, x_2)$, $h'(x)>0$, so $h(x)$ is increasing.
Note that $h(1)=0$, and when $x\to 0^+$, $h(x)\to +\infty$, when $x\to +\infty, h(x)\to -\infty$, therefore, the graph of $h(x)$ is illustrated as follows:
$h(x)$" />
Therefore, there is three zeros for $h(x)$, i.e. $x_3, x_4=1, x_5$.
In $(0,x_3)$ and $(1,x_5)$, $h(x)>0$, so $g(x)<0$, so $f'(x)<0$, so $f(x)$ is decreasing.
In $(x_3,1)$ and $(x_5,+\infty)$, $h(x)<0$, so $g(x)>0$, so $f'(x)>0$, so $f(x)$ is increasing.
Note that $f(1)=0$, and when $x\to 0^+$, $f(x)\to 1-2a\le 0$; when $x\to +\infty$, $f(x)\to 1-2a\le 0$. The graph of $f(x)$ can be illustrated as follows:
$f(x)$" />
Now, we can surely say, the maximum of $f(x)$ is $f(1)=0$, which means $a^x+a^{1/x}\le 2a$, with $\frac{1}{2}\le a<1$, $\forall x\in(0,+\infty)$.

Yes, it is true. Fix $a$ in the range $1/2 \le a < 1$. It suffices to prove
$$
a^x + a^{1/x}  \le 2a
$$
for $0 < x \le 1$, because the left-hand side is invariant under the substitution $x \leftrightarrow 1/x$. This inequality can be rearranged to
$$ 
g(x) := a^{-1/x} (2a - a^x) \ge 1 \,  .
$$
We have $g(1) = 1$, and we will now show that $g$ is decreasing on $(0, 1]$. The derivative is
$$
 g'(x) = \frac{a^{-1/x} \ln(a)}{x^2} \left( 2a - (x^2+1) a^x\right) \, .
$$
The first factor is negative, so we need to show that
$$
 2a - (x^2+1) a^x \ge 0 
$$
for $0 \le x \le 1$. It turns out to be convenient to set $y = 1-x$. With that substitution we need to show that
$$
 h(y) := 2a^y - 1- (1-y)^2 \ge 0
$$
for $0 \le y \le 1$. We have $h(0) = 0$ and $h(1) = 2a-1 \ge 0$, and the second derivative
$$
 h''(y) = 2 \left( \ln^2(a) \cdot a^y - 1\right)
$$
is negative. So $h$ is concave and therefore attains its minimum on the interval $[0, 1]$ at one of the boundary points.
We have thus shown that $h(y) \ge 0$, so that $g'(x) \le 0$, and therefore $g(x) \ge g(1) = 1$ for $0 < x \le 1$, and that completes the proof.
