Applying AM >= GM for two numbers 2^sinx and 2^cosx

(2^sinx + 2^cosx) / 2 >= (2^sinx.2^cosx)^1/2 

(2^sinx + 2^cosx) / 2 >= 2^{(sinx + cosx)/2}

now sinx + cosx = {root(2)}.sin(pi/4 + x) >= -root(2)

so (sinx + cosx) / 2 >= -1/root(2)

2^{(sinx + cosx)/2} >= 2^-1/root(2)

hence (2^sinx + 2^cosx) / 2 >= 2^-1/root(2)

2^sinx + 2^cosx >= 2(2^-1/root(2))

2^sinx + 2^cosx >= 2^{1 - 1/root(2)}

To prove    2^sinx+2^cosx > 2^(1-1/)

Let y = 2^sinx+2^cosx

Differentiating both the sides we obtain

dy/dx = Sin x . Cos x . 2^{sinx -1} - Cos x . Sin x . 2^{cosx - 1} 

For maxima or minima

dy/dx = 0

or  Sin x . Cos x . 2^{sinx -1} - Cos x . Sin x . 2^{cosx - 1}  = 0

or Sin x . Cos x . 2^{sinx -1} = Cos x . Sin x . 2^{cosx - 1} 

or 2^{sinx -1} =  2^{cosx - 1} 

or Sin x - 1= Cos x  - 1

or Sin x = Cos x

which is possible when x = pi / 4

or Sin x = Cos x  =1 / 

Thus for Sin x = Cos x  =1 / 

 y  =  2^sinx+2^cosx  = 2.2^1/

 This value is either maxima or minima, but second derivative test shows that it is minima,

(another way is that when x = 0, y = 3 which is greater than 2.2^1/  Hence later is minima)

Thus y  > or = 2.2^1/ > 2^(1-1/)

Hence Proved

Here I slightly edit the solution provided above

 To prove    y =2^sinx+2^cosx > 2^(1-1/)

 Differentiating both the sides we obtain

dy/dx = Sin x . Cos x . 2^{sinx -1} - Cos x . Sin x . 2^{cosx - 1} 

For maxima or minima

 dy/dx = 0

or  Sin x . Cos x . 2^{sinx -1} - Cos x . Sin x . 2^{cosx - 1}  = 0

or Sin x . Cos x . 2^{sinx -1} = Cos x . Sin x . 2^{cosx - 1} 

or 2^{sinx -1} =  2^{cosx - 1} 

or Sin x - 1= Cos x  - 1

or Sin x = Cos x

which is possible when x = pi / 4, 3 pi / 4 

Second derivative test reveals that 3pi/ 4 is the point of minima where y  min =  2^(1-1/)

 Also point of maxima is pi/ 4  and y max   =  2^sinx+2^cosx  = 2.2^1/

Thus y  > y min = 2^(1-1/)

or  2^sinx+2^cosx  >2^(1-1/)