Supremum and infimum of $(x + y)(1/x + 1/y)$

Question

I have $$A = \left\{\left(x + y\right)\left(x^{-1} + y^{-1}\right) \mid x,y>0 \right\}$$ and I'm looking for the supremum and infimum of this set. Rewriting the equation, I get $\dfrac{(x - y)^2}{xy} + 4$ so I conclude $\inf \text{A} = 4$.

My guess about the supremum is it doesn't exist. Is there a straightforward way to "see" this?

Answer 1

Yes, you are right, the supremum is equally to $+\infty$. To see this, you can just find a "convenient pair" of numbers that "reaches it" (approaches it). For this, fix $y=1$ and let $x\to \infty$. Then $$\frac{(x-y)^2}{xy}=\frac{x^2-2x+1}{x}=x-2+\frac1x\to+\infty$$ as $x\to+\infty$.

Answer 2

As for supremum, make $y =1$, and $(x-1)^2 /x$ tends to Infinity as $x$ tends to Infinity.

Answer 3

Another way to find the infimum is using the A.G.M. inequality. It has the advantage to generalise: for any $x_1,\dots,x_n>0$, $$(x_1+\dots+x_n)(x_1^{-1}+\dots+x_n^{-1})\ge n^2$$ Indeed, let's denote $A$, $G$ and $H$ the arithmetic, geometric and harmonic means of $x_1,\dots,x_n>0$. The inequality stipulates that $H\le G\le A$, with equality if and only if $x_1=\dots=x_n$.

Now $x_1+\dots+x_n=nA$, $\;x_1^{-1}+\dots+x_n^{-1}=\dfrac nH$, thus $$(x_1+\dots+x_n)(x_1^{-1}+\dots+x_n^{-1})=n^2=\frac AH\ge n^2.$$ Further this lower bound is attained for $x_1=\dots=x_n$. Thus it is the minimum of the product.