Let $a, b, u, v \in \mathbb{R}$ so that $av - bu = 1$.
Prove the following inequation:
$$a^2 + b^2 + u^2 + v^2 +au + bv \ge 3$$
The following is a proof for: $a^2 + b^2 + u^2 + v^2 +au + bv \ge 1$
$$\begin{align} a^2 + b^2 + u^2 + v^2 +au + bv \ge 1 \\\\ 2\frac{a^2 + b^2 + u^2 + v^2}{2} + 2\frac{au}{2} + 2\frac{bv}{2} \ge 1 \\\\ \frac{a^2 + 2au + u^2 + b^2 + 2bv + v^2}{2} + \frac{a^2 + b^2 + u^2 + v^2}{2} \ge 1 \\\\ \frac{(a + u)^2 + (b + v)^2}{2} + \frac{a^2 + b^2 + u^2 + v^2}{2} \ge 1 = av - bu \\\\ \frac{(a + u)^2 + (b + v)^2}{2} + \frac{a^2 + b^2 + u^2 + v^2}{2} - 2\frac{bu - av}{2} \ge 0 \\\\ \frac{(a + u)^2 + (b + v)^2}{2} + \frac{a^2 - 2av + v^2 + b^2 + 2bu + u^2}{2} \ge 0 \\\\ \frac{(a + u)^2 + (b + v)^2}{2} + \frac{(a - v)^2 + (b + u)^2}{2} \ge 0 \end{align}$$
The left part of the last inequation is a sum of nonnegative terms so it as a whole is nonnegative, or in other words greater than $0$.
Here's a proof for: $a^2 + b^2 + u^2 + v^2 \ge 2$
$$\begin{align} av - bu = 1 \\\\ 2bu - 2av = -2 \\\\ b^2 + 2bu + u^2 + a^2 - 2av + v^2 = a^2 + b^2 + u^2 + v^2 - 2 \\\\ (b + u)^2 + (a - v)^2 = a^2 + b^2 + u^2 + v^2 - 2 \ge 0 \\\\ (b + u)^2 + (a - v)^2 + 2 = a^2 + b^2 + u^2 + v^2 \ge 2 \end{align}$$
It's wrong.
Try $(a,b,u,v)=\left(1,0,\frac{1}{2},1\right).$
We need to prove that: $$1+\frac{1}{4}+1+\frac{1}{2}\geq3,$$ which is not true.
As said Will Jagy $$\min_{av-bu=1}(a^2+b^2+u^2+v^2+au+bv)=\sqrt3.$$
Indeed, we need to prove that: $$a^2+b^2+u^2+v^2+au+bv\geq\sqrt3(av-bu)$$ or $$a^2+(u-\sqrt3v)a+b^2+(v+\sqrt3u)+u^2+v^2\geq0,$$ for which it's enough to prove that $$(u-\sqrt3v)^2-4(b^2+(v+\sqrt3u)+u^2+v^2)\leq0$$ or $$4b^2+4(v+\sqrt3u)b+v^2+2\sqrt3uv+3u^2\geq0$$ or $$(2b+v+\sqrt3u)^2\geq0$$ and easy to see that the equality indeed occors.