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 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 $\sqrt{x^2 + y^2} \leq |x| + |y|$.
