Prove that $2^a+2^{1\over a} \le 1$ for $a<0$ 
QUESTION:
  Prove that if $a<0$, then $2^a+2^{1\over a} \le 1$.

How I started it:
So, if a<0, then there exists a number $u>0$ satisfying $a=-u$.
We have $2^{-u} + 2^{-1\over u} \le 1$ or by exponent rules
$1\over 2^u$$+$$1\over 2^{1\over u}$$\le 1$ but I got stuck here. Any help would be appreciated. Also, I need an algebraic non-calculus solution.
 A: set $$f(a)=2^a+2^{1\over a} = e^{a\ln2}+e^{\frac{1}{a}\ln2}\implies f'(a)=\ln2(2^a-\frac{2^{1\over a}}{a^2}) $$ 
it is not an obvious task to find the roots of $f'$ by hand 
despite that I opted for numerics. solving   $$f'(a)=\ln2(2^a-\frac{2^{1\over a}}{a^2}) =0~~~~~on~~~~ (-\infty, 0,)~~~ $$ gives  (see here) 
$$\color{red}{a_0\approx -4,64886,~~~  ~~~~a_1=-1~~~~and ~~~~ a_2\approx -0,215106}$$
Moreover, on $(-\infty, a_0,)\cup (a_1,a_2)~~~~$ $f'(a)<0$ hence $f$ decreases therein 
and on $ (a_0,a_1)\cup(a_2,0)~~~$ $f'(a)>0$ that is $f$ increases therein   see here
also,  $$\lim_{a\to 0^-} f(a)= 1=\lim_{a\to -\infty, }f(a)$$
Therefore it follows that on $(-\infty, 0)$ the function  $f$ has unique maximum at $a_1= -1$. Whereas, $a_0$ and $a_2$ are minima of $f$.
that is $$f(a)=2^a+2^{1\over a}\le f(-1)=1~~~~\forall ~~a<0$$
A: I solved it, but I didn't expect such a tedious solution. It is a proof by cases, first you assume that x is the one whose absolute value is greater than or equal to 1, while it's reciprocal's absolute value is less than or equal to 1 OR VICE VERSA, then I look at two cases, the one where $x\le -2$, and the one where $-2<x\le -1$. Using algebraic manipulations and Bernoulli's inequality you get to the answer.

A: Let $2^a=x$. 
Thus, $0<x<1$, $a=\log_2 x$ and we need to prove that
$$x+2^{\frac{1}{\log_2x}}\leq1$$ or
$$\frac{1}{\log_2x}\leq\log_2(1-x)$$ or
$$\log_2(1-x)\log_2x\leq1$$ or $f(x)\leq\ln^22,$
where $$f(x)=\ln(1-x)\ln{x}.$$
Indeed, $$f'(x)=-\frac{\ln{x}}{1-x}+\frac{\ln(1-x)}{x}=\frac{(1-x)\ln(1-x)-x\ln{x}}{x(1-x)}.$$
We'll prove that $f'(x)>0$ for $0<x<\frac{1}{2}$ and $f'(x)<0$ for $\frac{1}{2}<x<1.$
Let $g(x)=(1-x)\ln(1-x)-x\ln{x}.$
Hence, $$g'(x)=-\ln(1-x)-1-\ln{x}-1=\ln\frac{1}{e^2x(1-x)}$$
$g'(x)=0$ gives $x^2-x+\frac{1}{e^2}=0$, 
which gives that $g$ increases on $\left(0,\frac{1-\sqrt{1-\frac{4}{e^2}}}{2}\right]$ and on $\left[\frac{1+\sqrt{1-\frac{4}{e^2}}}{2},1\right)$
and it gives that $g$ decreases on $\left[\frac{1-\sqrt{1-\frac{4}{e^2}}}{2}, \frac{1-\sqrt{1+\frac{4}{e^2}}}{2}\right]$.
Now, since $$g\left(\frac{1}{2}\right)=\lim\limits_{x\rightarrow0^+}g(x)=\lim\limits_{x\rightarrow1^-}g(x)=0,$$
we see that $f'(x)$ changes the sign in $x=\frac{1}{2}$ from $+$ to $-$, which   ends the proof. 
