Prove that $2^a+2^{1\over a} \le 1$ for $a<0$

Question

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.

Answer 1

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 $$f(a)=2^a+2^{1\over a}\le f(-1)=1~~~~\forall ~~a<0$$

Answer 2

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.