Infimum of a set with two variables 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

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).
