I came upon the following question while working on an optimization problem:
Let $0 <\lambda < \frac{1}{2}$ be a parameter. Let $\phi:[0,1] \to [0,\lambda]$ be a smooth strictly increasing bijective function; in particular $\phi(0)=0,\phi(1)=\lambda$.
Question: Is it true that $$ A(\phi):=\int_0^1 (\phi'(r))^2r+\frac{(\phi(r))^2}{r} \ge \lambda^2$$ for every such $\phi$?
Note that equality holds for the homogeneous scaling $\phi(r)=\lambda r$.
I think that I have a proof for this claim when $\lambda \ge \frac{1}{2}$. However, this proof is really convoluted (so I won't reproduce it here), and I am really more interested in the regime $\lambda < \frac{1}{2}$.
Here is a proof for $A(\phi) \ge \frac{4}{5} \lambda^2$ (for all values of $\lambda$).
$$ A(\phi) =\int_0^1 (\phi' \sqrt{r})^2+(\frac{\phi}{\sqrt r})^2 \ge (\int \phi' \sqrt{r} )^2+(\int \frac{\phi}{\sqrt r} )^2,$$
where we have used the C-S inequality. Now, integration by parts gives
$\int \frac{\phi}{\sqrt r}=-2\int \phi' \sqrt{r}+2\lambda$, so setting $I:=\int \phi' \sqrt{r}$, we obtain that that $$ A(\phi) \ge I^2+(-2I+2\lambda)^2=5I^2-8\lambda I+4\lambda^2 \ge c \lambda^2$$
if and only if $5I^2-8\lambda I+(4-c)\lambda^2 \ge 0$. Thinking of $\lambda, c$ as constant parameters, we want the quadratic in $I$ to never vanish; this happens exactly when $64\lambda^2-20(4-c)\lambda^2 \le 0$. This holds iff $64 \le 20(4-c) \iff 16 \le 5(4-c) \iff c \le \frac{4}{5}$.
Important remark: After using the C-S inequality in this proof, we cannot expect a better constant than $\frac{8}{9}$, which is what we get in $(\int \phi' \sqrt{r} )^2+(\int \frac{\phi}{\sqrt r} )^2$ when we plug in $\phi(r)=\lambda r$.
Since $a^2 + b^2 \ge 2ab$, we have $$ \int_0^1 \phi'(r)^2r + \frac{\phi(r)^2}{r} \,dr \ge \int_0^1 2\phi(r)\phi'(r)\,dr = \int_0^1 \frac{d}{dr}(\phi^2(r))\,dr = \phi^2(1)-\phi^2(0) = \lambda^2. $$