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Question Number 41957 by behi83417@gmail.com last updated on 15/Aug/18

Commented by MJS last updated on 16/Aug/18

$I {don}^{'} t see any solution with all variables \in R$ $in C we can freely choose 3 variables and solve$ $for the remaining, so {there}^{'} s no unique solution$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18

	$\frac{a x + b y}{2} ⩾ \sqrt{a x b y}$ $\frac{a y + b x}{2} ⩾ \sqrt{a y b x}$ $(a x + b y) ((a y + b x) ⩾ 4 a b x y$ $s o$ $(x^{2} + y^{2}) (a^{2} + b^{2}) ⩾ 4 a b x y$ $x^{2} a^{2} + x^{2} b^{2} + a^{2} y^{2} + b^{2} y^{2} - 4 a b x y ⩾ 0$ ${(a x - b y)}^{2} + ({(a y - b x)}^{2} ⩾ 0$ $i f c o n s i d e r$ ${(a x - b y)}^{2} + {(a y - b x)}^{2} = 0$ $t h e n a x = b y$ $\frac{a}{y} = \frac{b}{x} = r$ $a = r y b = r x$ $a y = b x$ ${r y}^{2} = {r x}^{2}$ $r (x^{2} - y^{2}) = 0$ $s o x = y x = y$ $h e n c e a = r y$ $b = r x = r y$ $s o a = b a = b$ $(a x + b y) (a y + b x) = (x^{2} + y^{2}) (a^{2} + b^{2})$ $(a x + a x) (a x + a x) = (x^{2} + x^{2}) (a^{2} + a^{2})$ $4 a^{2} x^{2} = 4 x^{2} a^{2}$ $c o r e l a t i o n b e t w e e n a b a n d x y c a n n o t b e e s t a b l i s h e d$ $t o f i n d \frac{a b}{x y}$
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