show that $(0.5)^x+(0.5)^{1/x}\le 1.$ if $x>0$ let $x>0$ show that
$$(0.5)^x+(0.5)^{1/x}\le 1.x>0,\tag{1}$$
Here is what I tried:let $f(x)=(0.5)^x+(0.5)^{1/x}$ so $f(x)=f(\frac{1}{x})$.si we only prove when $x\ge 1$.have $f(x)\le 1=f(1)$.it is sufficient to prove that
$$f'(x)=(0.5)^x\ln{0.5}-\dfrac{1}{x^2}\cdot(0.5)^{\frac{1}{x}}\ln{0.5}\le 0,x\ge 1$$
since $\ln{0.5}<0$,if we can show that
$$(0.5)^{\frac{1}{x}-x}\le x^2,x\ge 1$$
if we let $g(x)=x^2-(0.5)^{\frac{1}{x}-x}$,then $$g’(x)=2x+\ln{0.5}(0.5)^{\frac{1}{x}-x}\left(\dfrac{1}{x^2}+1\right)$$
it seem not easy to prove $g'(x)\ge 0,x\ge 1$ because $\ln{0.5}<0$.other words,this methods following more ugly!,so  How to prove this inequality $(1)$.Thanks.
 A: We can make the OP's approach work. Let $f(x) = (0.5)^x+(0.5)^{1/x}$, so that $f(x) = f(\frac1x)$, and it suffices to prove that $f(x)<1$ for $0<x<1$. We will do so by showing that $f(x)$ is decreasing to a local minimum and then increasing on this interval.
Since $f'(x) = (2^{-1/x}x^{-2}-2^{-x}) \ln 2 = (2^{x-1/x}-x^2)\frac{\ln 2}{x^22^x}$, we see that $f(x)$ is decreasing when $x^2 > 2^{x-1/x}$ and increasing otherwise. Equivalently (and this is, perhaps, the crucial step), $f(x)$ is decreasing when $2\ln x > (x-\frac1x)\ln 2$ and increasing otherwise. If we set $h(x) = 2\ln x - (x-\frac1x)\ln 2$, we therefore want to show that $h(x)$ starts off negative, has a unique zero, and is then positive over the interval $(0,1)$.
But now $h'(x) = \frac2x - (1+\frac1{x^2})\ln 2 = -(\ln2)(x^2-\frac2{\ln2}x+1)/x^2$; this makes it easy to see that $h'(x)>0$ precisely when
$$
0.40281 \approx \frac{1-\sqrt{1-(\ln2)^2}}{\ln2} < x < \frac{1+\sqrt{1-(\ln2)^2}}{\ln2} \approx 2.48258,
$$
which concludes the proof.
A: An alternative approach: Substitute $2^{-x} = u$. Then $0 < u < 1$ and $ 2^{-1/x} = 2^{\ln 2 /\ln u}$ so that
$$
\begin{align}
 &\, 2^{-x} +  2^{-1/x} \le 1 \\
\iff &\, u + 2^{\ln 2 /\ln u} \le 1 \\
\iff &\, \frac{(\ln 2)^2}{\ln u} \le \ln (1-u) \\
\iff &\, \ln(u) \ln (1-u) \le (\ln 2)^2 \, .
\end{align}
$$
Therefore we consider the function $g(u) = \ln(u) \ln (1-u) $ on the interval $(0, 1)$. $g$ is symmetric with respect to $u=1/2$, so that $g'(1/2) = 0$. The idea is now to show that $g$ is concave. Then it follows that $g$ has a maximum at $u=1/2$, i.e.
$$
\ln(u) \ln (1-u) = g(u) \le g(\frac 12) = (\ln 2)^2
$$
and we are done.
A straightforward calculation shows that the second derivative of $g$ is
$$
 g''(u) = -\frac{\log(1-u)}{u^2} -\frac{\log(u)}{(1-u)^2} - \frac{2}{u(1-u)} \, .
$$
Using the well-known inequality $\ln(y) \le y-1$ we have
$$ 
-\log(1-u) = \log \frac{1}{1-u} \le  \frac{u}{1-u}
$$
and
$$ 
-\log(u) = \log \frac{1}{u} \le  \frac{1-u}{u} \, .
$$
Substituting these estimates in $g''(u)$ gives $g''(u) \le 0$. So $g$ is concave on $(0, 1)$ and that concludes the proof.
