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proof that infimum $\inf(\sqrt{n} +1/\sqrt{n}: n \in\mathbb{N})= 2.$

firstly i checked if its a lower bound of this set

$\sqrt{n} +1/\sqrt{n}\geq 2/\sqrt{n}$

$n+1\geq 2\sqrt{n}$

$n-2\sqrt{n}+1\geq 0$

$n^2-2n+1\geq 0$

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

which is always true

now i wanted to show that $2$ is not only a lower bound but also the greatest of them all and i got a little bit stuck

so I let some $\epsilon>0$ and i know that it should exist some $n_0$ that will prove this

$\sqrt{n} +1/\sqrt{n}<2+\epsilon$

$\sqrt{n} +1/\sqrt{n}-2<\epsilon$

$\sqrt{n}(1+1/n)-2<\epsilon$

and at this point i tried a lot of things but i dont know if its going the right way

Nik
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Karol Bargieł
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1 Answers1

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For each $n\in\Bbb N$,\begin{align}\sqrt n+\frac1{\sqrt n}\geqslant2&\iff\sqrt n-2+\frac1{\sqrt n}\geqslant0\\&\iff\left(\sqrt[4]n-\frac1{\sqrt[4]n}\right)^2\geqslant0,\end{align}and, since this inequality holds, you have indeed $\sqrt n+\frac1{\sqrt n}\geqslant2$. But$$2\in\left\{\sqrt n+\frac1{\sqrt n}\,\middle|\,n\in\Bbb N\right\},$$since $2=\sqrt1+\frac1{\sqrt1}$. Since $2$ belongs to the given set and it is also a lower bound, it is the greatest lower bound.

José Carlos Santos
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