show that inequality (2raised to x)+(2raised to y)>or=3 is true when x,y>0 and x^2+y^2=1?

The condition of the problem,`x^2+y^2=1` , has inspired
the next approach.

Put x = sin t and y = cos t and tan
(`t/2` ) = z.

`sin t = (2tan (t/2))/(1 + tan^2 (t/2)) lt=gt
x = 2z/(1+z^2)`

```cost`  = `(1-tan^2 (t/2))/(1 + tan^2
(t/2)) lt=gty = (1-z^2)/(1+z^2)`

Using the mean inequality,
you may write:

`2^x + 2^y = 2^x + 2^(y-1) + 2^(y-1) gt=
3*2^((x+2y-2)/3)gt=` 3 <=> x+2y-2 `gt=`
0

Substitute x and y in the inequality x+2y-2 `gt=`
0:

`(2z)/(1+z^2) + (2-2z^2)/(1+z^2) - 2 gt=` 
0

`2z + 2 - 2z^2 - 2 - 2z^2gt= `
0

Add similar members and remove opposite
terms:

`-4z^2+ 2z gt=` 
0

Multiply by -1:

`4z^2 - 2z
=lt`  0

Divide by 2 => `2z^2 - z=lt`  0 =>
z(2z-1)`=lt` 0 => z `in`  [0;1/2] (a)

Using the mean
inequality, you may write:

`2^x + 2^y = 2^(x-1) + 2^(x-1) +
2^y gt=`` 3*2^((2x-2+y)/3)gt=` 3 <=> 2x-2+y`gt=`
0

Substitute x and y in the inequality 2x-2+y`gt=`
0:

`(4z)/(1+z^2) - 2 + (1-z^2)/(1+z^2) gt=` 
0

`4z + 1 - z^2 - 2 - 2z^2`  `gt=` 
0

`- 3z^2 + 4z - 1 gt= 0`  => `3z^2- 4z+ 1 =lt 0` 
<=> z `in ` [`1/3;1` ]
(b)

ANSWER: Relations (a) and (b) prove `2^x
+ 2^y gt= 3` , if `x^2 + y^2 = 1`