Short distance teleportation

Algebra Level 5

{ 1 1 + 2 x 2 + 1 1 + 2 y 2 = 2 1 + 2 x y x x ( 1 2 x ) + y ( 1 2 y ) = 119 3200 39 2 \begin{cases} \dfrac{1}{\sqrt{1+2x^{2}}}+\dfrac{1}{\sqrt{1+2y^{2}}}=\dfrac{2}{\sqrt{1+2xy}} \\ x\sqrt{x(1-2x)}+\sqrt{y(1-2y)}=\dfrac{119}{3200} \cdot \sqrt{\dfrac{39}{2}} \end{cases}

If x x and y y satisfy the system of equations above, find the maximum value of x + y x+y .


The answer is 0.975.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The question gives the conditions: 1 + 2 x y > 0 1+2xy > 0 , x ( 1 2 x ) 0 x(1-2x)\geq0 and y ( 1 2 y ) 0 y(1-2y)\geq0 . By solving them, it is obtained that 0 x 1 2 0\leq x\leq\dfrac{1}{2} and 0 y 1 2 0\leq y\leq\dfrac{1}{2} .

Below, we will show that for 0 x , y 1 2 0\leq x,y\leq\dfrac{1}{2} , 1 1 + 2 x 2 + 1 1 + 2 y 2 2 1 + 2 x y \dfrac{1}{\sqrt{1+2x^{2}}}+\dfrac{1}{\sqrt{1+2y^{2}}}\leq\dfrac{2}{\sqrt{1+2xy}} 1 1 + 2 x 2 + 1 1 + 2 y 2 + 2 ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) 4 1 + 2 x y \Leftrightarrow\dfrac{1}{1+2x^{2}}+\dfrac{1}{1+2y^{2}}+\dfrac{2}{\sqrt{(1+2x^{2})(1+2y^{2})}}\leq\dfrac{4}{1+2xy}

( 1 1 + 2 x 2 + 1 1 + 2 y 2 2 1 + 2 x y ) + ( 2 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 1 + 2 x y ) 0 \Leftrightarrow\left(\dfrac{1}{1+2x^{2}}+\dfrac{1}{1+2y^{2}}-\dfrac{2}{1+2xy}\right)+ \left(\dfrac{2}{\sqrt{4x^{2}y^{2}+2x^{2}+2y^2+1}}-\dfrac{2}{1+2xy}\right)\leq0

2 ( x y ) 2 ( 1 2 x y ) ( 1 + 2 x y ) ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) + ( 2 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 1 + 2 x y ) 0 \Leftrightarrow-\dfrac{2{(x-y)}^{2}(1-2xy)}{(1+2xy)(1+2x^2)(1+2y^2)}+ \left(\dfrac{2}{\sqrt{4x^{2}y^{2}+2x^{2}+2y^2+1}}-\dfrac{2}{1+2xy}\right)\leq0

We know that 2 ( x y ) 2 ( 1 2 x y ) ( 1 + 2 x y ) ( 1 + 2 x 2 ) ( 1 + 2 y 2 ) 0 \dfrac{2{(x-y)}^{2}(1-2xy)}{(1+2xy)(1+2x^2)(1+2y^2)}\leq0 So we have to prove that the second term on the left hand side is less than 0. 2 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 1 + 2 x y \dfrac{2}{\sqrt{4x^{2}y^{2}+2x^{2}+2y^2+1}}\leq\dfrac{2}{1+2xy}

2 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 2 4 x 2 y 2 + 4 x y + 1 \Leftrightarrow \dfrac{2}{\sqrt{4x^{2}y^{2}+2x^{2}+2y^2+1}}\leq\dfrac{2}{\sqrt{4x^{2}y^{2}+4xy+1}} This implies that 4 x 2 y 2 + 2 x 2 + 2 y 2 + 1 4 x 2 y 2 + 4 x y + 1 4x^{2}y^{2}+2x^{2}+2y^2+1\geq 4x^{2}y^{2}+4xy+1 , which is obviously true. Hence, our claim is true, with equality holding when x = y x=y . Substituting this into the second equation and solving, we get the maximum values to be x = y = 0.4875 x=y=0.4875 Hence, the solution is thus found.

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...