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

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### 1 Answer

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` **