Let $\lambda>1$ be a parameter. Let $\psi:[0,1] \to [0,\lambda]$ be a smooth surjective function, satisfying $\psi(0)=0$.
Question: Is it true that $$ E(\psi):=\int_0^1 \big((\psi'(r)-1)^2+(\frac{\psi(r)}{r}-1)^2\big) rdr \ge (1-\lambda)^2 \, \, \,?$$
I prove below that this is true if $\psi(1)=\lambda$. I am asking about what happens without this assumption.
Note that for $\psi(r)=\lambda r$ equality clearly holds.
Proof:
We shall use $a^2 + b^2 \ge 2ab$ with $a=\psi'(r)-1, b=\frac{\psi(r)}{r}-1$: $$ \begin{split} & E(\psi) \ge \int_0^1 2(\psi'(r)-1)(\frac{\psi(r)}{r}-1) rdr = \\ & \int_0^1\frac{d}{dr} \big((\psi(r)-r)^2\big) dr=(\psi(1)-1)^2. \end{split} $$
Since we assumed $\psi(1)=\lambda$, we are done.
Replacing $\int_0^1 $ by $\int_0^a$ in the argument above, we get $ E(\psi) \ge (\psi(a)-a)^2$, thus $ E(\psi) \ge\max_{a \in [0,1]} (\psi(a)-a)^2$.
Comment:
The statement holds for $0 <\lambda \le 1$. Indeed, in that case $1-\psi(1) \ge 1-\lambda \ge 0$, so $E(\psi) \ge (1-\psi(1))^2 \ge (1-\lambda)^2.$
Thus, the interesting part is for $\lambda > 1$, as assumed in the question.
The answer is positive.
As proved in the question, $E(\psi) \ge (\psi(a)-a)^2$ for any $a\in [0,1]$. In particular, take $a$ satisfying $\psi(a)=\lambda$. Then, since $a \le 1 < \lambda$, we have $\lambda-a \ge \lambda-1>0$, so
$$ E(\psi) \ge (\lambda-a)^2 \ge (\lambda-1)^2. $$