Infimum of a set with two variables

Question

I have encountered a problem concerning the infimum of a set:

Prove that. $$\mathrm {inf} \left\lbrace\sqrt{a^2+{1\over b^2}}:a,b\in(0,1) \right\rbrace=1$$

What I've been able to do is to prove that: $$\mathrm{inf}\left\lbrace a^2:a\in(0,1) \right\rbrace=0$$ and $$\mathrm{inf}\left\lbrace {1\over{b^2}}:b\in(0,1) \right\rbrace=1$$ combining these two results bearing in mind that $\sqrt1=1$ I've concluded that the above equation holds. However my approach doesn't seem rigorous enough. Could someone give me hint or an advice on how to refine my proof.

Thanks

Answer 1

You cannot just take the infimum over $a$ and $b$ separately. The correct proof goes something like this:

Since $a>0$, $0<b<1$, we have, $$\sqrt{a^2+\frac{1}{b^2}}> \frac{1}{b} > 1$$

so the infimum is $\ge 1$. Now you are done, if you can choose a sequence $a_n,b_n\in (0,1)$ such that $\sqrt{a_n^2+\frac{1}{b_n^2}}\rightarrow 1$ (why? and do that).

Answer 2

You can't just take the inf over 2 variables separately, but the fact they can be chosen independently means this is good working.

To show that $1$ is the greatest lower bound. You can see it's a lower bound as that quantity always bigger than $1$.

Then to show it's the biggest, you can show that for any $\epsilon > 0$, $1 + \epsilon$ isn't a lower bound (any greater lower bound could be written in this form). You can do this choosing $0<a<1$ really close to $0$ and $0<b<1$ really close to 1 to get $\sqrt{a^2 + 1/b^2}$ less than $1+\epsilon$. You may find it useful to note that $\sqrt{x^2 + y^2} \leq |x| + |y|$.

Answer 3

For example, and since $\;0<a,b<1\;$ (*)

$$\sqrt{a^2+\frac1{b^2}}=\frac1b\sqrt{a^2b^2+1}\ge 1\iff a^2b^2+1\ge b^2\iff b^2(1-a^2)\le 1$$

and the last inquality follows from (*), as $\;0<b^2\,,\,1-a^2<1\;$.

On the other hand, whenever $\;a\to 0\;$ and $\;b\to 1\;$ , we get that $\;\sqrt{a^2+\frac1{b^2}}\to 1\;$