<s>[INST] $xy=1 \implies $minimum $x+y=$? If $x,y$ are real positive numbers such that $xy=1$, how can I find the minimum for $x+y$?
 [/INST] as $xy=1$;
so, $y=1/x$;
as we know that 
$(x+y)^2 = x^2+y^2+2xy$;
$(x+y)^2 = x^2 = 1/x^2 +2$;
so, $x+y = \sqrt{x^2 + \frac{1}{x^2} + 2}$
to be minimum, compare RHS by 0;
you will get $x=-1$; but as $x$ is a +ve real number, 
so, $x$ is minimum at 1;
so $x+y =2$;
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