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

Saturday, January 11, 2014

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, $\displaystyle{{x}}^{{2}}+{{y}}^{{2}}={1}$ , has inspired
the next approach.

Put x = sin t and y = cos t and tan
( $\displaystyle\frac{{t}}{{2}}$ ) = z.

$\displaystyle{\sin{{t}}}=\frac{{{2}{\tan{{\left(\frac{{t}}{{2}}\right)}}}}}{{{1}+{{\tan}}^{{2}}{\left(\frac{{t}}{{2}}\right)}}}\leq\gt$
x = 2z/(1+z^2)

$\displaystyle{\cos{{t}}}$ = $\displaystyle\frac{{{1}-{{\tan}}^{{2}}{\left(\frac{{t}}{{2}}\right)}}}{{{1}+{{\tan}}^{{2}}}}$
(t/2)) lt=gty = (1-z^2)/(1+z^2)

Using the mean inequality,
you may write:

$\displaystyle{{2}}^{{x}}+{{2}}^{{y}}={{2}}^{{x}}+{{2}}^{{{y}-{1}}}+{{2}}^{{{y}-{1}}}\geq$
3*2^((x+2y-2)/3)gt= $\displaystyle{3}\Leftrightarrow{x}+{2}{y}-{2}$ gt=
0

Substitute x and y in the inequality x+2y-2 $\displaystyle\geq$
0:

$\displaystyle\frac{{{2}{z}}}{{{1}+{{z}}^{{2}}}}+\frac{{{2}-{2}{{z}}^{{2}}}}{{{1}+{{z}}^{{2}}}}-{2}\geq$
0

$\displaystyle{2}{z}+{2}-{2}{{z}}^{{2}}-{2}-{2}{{z}}^{{2}}\geq$
0

Add similar members and remove opposite
terms:

$\displaystyle-{4}{{z}}^{{2}}+{2}{z}\geq$
0

Multiply by -1:

$\displaystyle{4}{{z}}^{{2}}-{2}{z}$
=lt $\displaystyle{0}$

Divide by 2 => $\displaystyle{2}{{z}}^{{2}}-{z}=\lt$ 0 =>
z(2z-1) $\displaystyle=\lt$ 0 => z $\displaystyle\in$ [0;1/2] (a)

Using the mean
inequality, you may write:

$\displaystyle{{2}}^{{x}}+{{2}}^{{y}}={{2}}^{{{x}-{1}}}+{{2}}^{{{x}-{1}}}+$
2^y gt= 3*2^((2x-2+y)/3)gt= $\displaystyle{3}\Leftrightarrow{2}{x}-{2}+{y}$ gt=
0

Substitute x and y in the inequality 2x-2+y $\displaystyle\geq$
0:

$\displaystyle\frac{{{4}{z}}}{{{1}+{{z}}^{{2}}}}-{2}+\frac{{{1}-{{z}}^{{2}}}}{{{1}+{{z}}^{{2}}}}\geq$
0

$\displaystyle{4}{z}+{1}-{{z}}^{{2}}-{2}-{2}{{z}}^{{2}}$ $\displaystyle\geq$
0

$\displaystyle-{3}{{z}}^{{2}}+{4}{z}-{1}\geq{0}$ => $\displaystyle{3}{{z}}^{{2}}-{4}{z}+{1}=\lt{0}$
<=> z $\displaystyle\in$ [ $\displaystyle\frac{{1}}{{3}};{1}$ ]
(b)

ANSWER: Relations (a) and (b) prove $\displaystyle{{2}}^{{x}}$
+ 2^y gt= 3 $\displaystyle,{\quad\text{if}\quad}$ x^2 + y^2 = 1

Help Notes

Saturday, January 11, 2014