Given positive $x,y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $, find minimum $(x+y)$

Question

I am given positive numbers $x, y$ such that $x > y$ and $\sqrt{x} \sqrt{y}(x-y) = x+y $. I need to find the minimum value of $(x+y)$. Here is my try. Using AM-GM inequality for nonnegative numbers, I have

$$ \frac{(x+y)}{2} \geqslant \sqrt{x} \sqrt{y} $$

$$ \sqrt{x} \sqrt{y}(x-y) \geqslant 2 \sqrt{x} \sqrt{y} \\ \therefore (x-y) \geqslant 2 $$

So, I have been able to arrive at this conclusion. But I am stuck here. Any help ?

Thanks

Answer 1

By AM-GM $$(x+y)^2=xy(x-y)^2=\frac{1}{4}\cdot4xy(x-y)^2\leq\frac{1}{4}\left(\frac{4xy+(x-y)^2}{2}\right)^2=\frac{(x+y)^4}{16},$$ which gives $$x+y\geq4.$$ The equality occurs for $(x-y)\sqrt{xy}=x+y$ and $4xy=(x-y)^2,$ which gives $$(x,y)=(2+\sqrt2,2-\sqrt2),$$ which says that we got a minimal value.

Answer 2

put $x=r^2{cos}^2a$ and $y=r^2{sin}^2a$ also let $a$ belong to $[0,\frac{\pi}{2}]$

thus we have to find max value of $r^2$

plugging the values in the given equation and simplifying using basic trig formulas we have $r^4(cosa)(sina)(cos2a)=r^2$ or

$ r^2=\frac{4}{sin(4a)} \ge 4$

Answer 3

Hint : put $x=\alpha \cosh^2(x)$ and $y=\alpha\sinh^2(x)$ the condition becomes :

$$\alpha=\tanh(x)+\frac{1}{\tanh(x)}$$

The expression is :

$$x+y=\Big(\tanh(x)+\frac{1}{\tanh(x)}\Big)\frac{1+\tanh^2(x)}{1-\tanh^2(x)}$$

Solving it we found $x+y\geq 4$.

Answer 4

Given $\sqrt{xy}\left(x-y\right)=x+y$

let $yx=c$ , where $c>0$.

$$\sqrt{c}\left(x^{2}-c\right)=x^{2}+c$$ $$x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -[1]$$

Let a function $$F(x,c)=x^{2}-\frac{\left(c+1+2\sqrt{c}\right)}{\left(c-1\right)}c$$ be defined. Then $$\frac{\partial F(x,c)}{\partial c}=0$$ at a constant $x$ gives us $c \approx 2.618 \implies x \approx 3.33 $ ( using $[1]$ ) . So, $$x+\frac{c}{x}\geqslant 4$$ $$min(x+y)=4$$

when $x=2+\sqrt{2} \text{ and } y=2-\sqrt2$ as pointed out.

Answer 5

Hint.

Making

$$ \cases{ u = x+y\\ v = x-y } $$

we have

$$ \sqrt{u^2-v^2}=2\frac uv $$

so

$$ u^2 = \frac{v^4}{v^2-4} $$

etc.