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

Re:Prove this

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