solution of equation

2^{x} + 2^{\left|x \right|}\geq 2\sqrt{2}

\text{When } x\geq 0,

2^{x+1}\geq 2^{3/2}

\Rightarrow x+1\geq \frac{3}{2} \Rightarrow x\geq \frac{1}{2}

\text{When } x<0,

2^{x}+2^{-x}\geq 2^{3/2}

\text{Take }2^{x}\text{ = t,}

t^{2}-2\sqrt{2}t +1\geq 0

\Rightarrow 8t^{2}-4\leq 0\Rightarrow t^{2}\leq 1/2

\Rightarrow \frac{-1}{\sqrt{2}}\leq t\leq \frac{1}{\sqrt{2}}

\text{But }2^{x}\geq 0

2^{x}\leq 2^{-1/2}\Rightarrow x\leq -1/2

\boxed{x\in(-\infty,-\frac{1}{2}]\cup[\frac{1}{2},\infty)}

ashish u hv made a mistake in ur 2nd portion
when x<0

2^x +2^-x >=2√2
let 2^x =t

t+1/t >= 2√2
or, t² - 2√2t +1 >=0
or t>=√2 -1
or 2^x >√2 -1
taking log, we have,

x<log₂(√2 -1) ≈ 1.27

You're correct, I messed up this one totally. [3]

t^{2}-2\sqrt{2}t+1\geq 0

\Rightarrow \left( t-\sqrt{2}\right)^{2}\geq 1

\Rightarrow \left(t-\sqrt{2} \right)\geq 1 \text{ or } \left(t-\sqrt{2} \right)\leq -1

\Rightarrow t\leq \sqrt{2}-1 \text{ or } t\geq \sqrt{2}+1

\Rightarrow 2^{x}\leq \sqrt{2}-1 \text{ or } 2^{x}\geq \sqrt{2}+1

\Rightarrow x\leq \log_{2} \left(\sqrt{2}- 1\right)

x \in \left( -\infty,\log_{2} \left(\sqrt{2}-1\right)\right] \cup \left[ \frac{1}{2},\infty\right)

I guess this should be the answer. That is if I'm not making some other stupid mistake again! :P