Minimum value #2

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

$\Rightarrow xy(x - y)^2 = (x + y)^2$

$\Rightarrow xy((x + y)^2 - 4xy) = (x + y)^2$

$(x + y)^2(xy - 1) = 4x^2y^2$

$\Rightarrow (x + y)^2 = \dfrac{4x^2y^2}{xy - 1}$

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$(xy - 2)^2 \geq 0$

$x^2y^2 \geq 4(xy - 1)$

$\Rightarrow \dfrac{x^2y^2}{xy - 1} \geq 4$

$\Rightarrow (x + y)^2 = \dfrac{4x^2y^2}{xy - 1} \geq 4\cdot 4 = 16$

$\Rightarrow x + y \geq \boxed{4}$

Equality holds when $xy = 2 , x + y = 4 \Rightarrow \boxed{x = 2 + \sqrt{2} , y = 2 - \sqrt{2}}$

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

$\Longrightarrow x>y>0$ and

${ \left( x+y \right) }^{ 2 }=xy{ \left( x-y \right) }^{ 2 }=\frac { 1 }{ 4 } 4xy\left[ { \left( x+y \right) }^{ 2 }-4xy \right] \\ { \left( x+y \right) }^{ 2 }\le \frac { 1 }{ 4 } { \left[ \frac { 4xy+{ \left( x+y \right) }^{ 2 }-4xy }{ 2 } \right] }^{ 2 }\left( AM-GM\quad inequality, \quad substitute \quad a=4xy,b={ \left( x+y \right) }^{ 2 }-4xy \right) \\ { \left( x+y \right) }^{ 2 }\le \frac { 1 }{ 16 } { \left( x+y \right) }^{ 4 }\\ \Longrightarrow E=x+y\ge 4$

$E=4$ when

$\begin{cases} { \left( x+y \right) }^{ 2 }-4xy=4xy \\ x+y=4 \end{cases}\Longleftrightarrow \begin{cases} xy=2 \\ x+y=4 \end{cases}\Longrightarrow \begin{cases} x=2+\sqrt { 2 } \\ y=2-\sqrt { 2 } \end{cases}$

Hence, ${ E }_{ min }=\boxed { 4 } .$

Note $x+y=s$ and $xy=p$ . Then ${ (x-y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-2xy={ s }^{ 2 }-4p$ so $x-y=\sqrt { { s }^{ 2 }-4p }$ . The equality is equivalent to:
$\sqrt { p } \cdot \sqrt { { s }^{ 2 }-4p } =s\ <=> p({ s }^{ 2 }-4p)={ s }^{ 2 } <=> { s }^{ 2 }p=4{ p }^{ 2 }+{ s }^{ 2 }$ By AM-GM we have $4{ p }^{ 2 }+{ s }^{ 2 }\ge 4ps$ ( )
Then ${ s }^{ 2 }p\ge 4ps<=>s\ge 4$
So the minimum value is attained when we have equality in (

), so $2p=s=4$ .
To find the values of $x$ and $y$ we need to solve the system
$\begin{cases} xy=2 \\ x+y=4 \end{cases}$ which gives the unique solution $x=2+\sqrt { 2 }$ and $y=2-\sqrt { 2 }$ .
Hence the minimum value of $x+y$ is 4 .

$u = x + y$

$v = x - y$

$x = \frac {u + v} {2}$

$y = \frac {u - v} {2}$

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

$v^2 (u^2 - v^2) = 4 u^2$

$u^2(v^2 - 4) = v^4$

$u = \frac {v^2} {\sqrt {v^2 - 4}} = \sqrt {v^2 - 4} + \frac {4} {\sqrt {v^2 - 4}} \geq \boxed {4}$ ( AM - GM )

Equality holds when $\sqrt {v^2 - 4} = \frac {4} {\sqrt {v^2 - 4}} \implies v^2 - 4 = 4 \implies v = 2 \sqrt {2}$

$x = 2 + \sqrt {2}$ , $y = 2 - \sqrt {2}$

Here is an alternative type of solution without using AM - GM (using a single, newly introduced variable and differentiation instead):

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

$x, y \in \mathbb {R}^+$

It is easy to see, that x > y (otherwise LHS ≤ 0 while RHS > 0, which would be a contradiction).

$\text {Let } a = \frac {y}{x} \Rightarrow y = ax , a \in \mathbb {R}^+ , 0 < a < 1$

$\sqrt {ax^2} (1 - a)x = (a + 1)x$

$x = \frac {a + 1}{ \sqrt {a} (1 - a) }$

$y = ax = \frac { \sqrt {a} (a + 1) }{1 - a}$

$f(a) = x + y = \frac { (a + 1)^2 }{ \sqrt {a} (1 - a) }$

Let's find the stationary points of f(a) ( f'(a) = 0, using the quotient and chain rules):

$\frac {2(a + 1) \sqrt {a} (1 -a) - (a + 1)^2 ( 0.5 ÷ \sqrt {a} - 1.5 \sqrt {a} ) }{ ( \sqrt {a} (1 - a) )^2 }$ = 0

After multiplying/dividing by the non-zero factors (0 < a < 1) and multiplying by 2, we get:

$4(1 - a) - (a + 1)( \frac {1}{a} - 3 ) = 0$

After expanding, simplifying and (× (-a) ) :

$a^2 - 6a + 1 = 0$

$a = 3 ± 2 \sqrt {2}$

$\text {Since 0 < a < 1, the only solution is: } a = 3 - 2 \sqrt {2}$

(For the sake of completeness, we have to see whether it is really a minimum by either determining the sign of the second derivative of f(a) or the sign change of the first derivative about our stationary point. They both confirm the minimum in this case.)

This gives us:

$x= 2 + \sqrt {2} , y = 2 - \sqrt {2} \text { and } x + y = \boxed {4}$