How to solve this inequality $3> 2^x+2^{1-x}$

Question

By hit and trial I get the answer as $x\in\mathbb(0,1)$ but I don't think about the actual process by which this inequality can solve. Plz tell any process by which I solve this ? I take log both sides , but this can't make any sense.

Answer 1

It's $$2^{2x}-3\cdot2^x+2<0$$ or $$(2^x-2)(2^x-1)<0$$ or $$1<2^x<2$$ or $$0<x<1.$$

Answer 2

Hints:

• Set $y = 2^x \Rightarrow 3 > y + \frac{2}{y}$.
• Solve $3 =y + \frac{2}{y}$.
• Take $3$ test values (to the left, between and to the right of the roots) and reason via continuity.

Answer 3

Another way is to note that $2^t$ is convex and we have $x\in (0,1)\implies (1,0)\succ (x,1-x)$ while $x\not \in (0,1)\implies (x,1-x)\succ (1,0)$. Then Karamata’s inequality ( https://en.m.wikipedia.org/wiki/Karamata%27s_inequality) directly gives the solution set is $x\in (0,1)$