Minimum value of PA+PB+AB

Question

If $P(2,1) $ and $A $ and $B $ lie on $x$ axis and $y=x $ respectively, then find the minimum value of $PA+PB+AB $ .

If $A$ was given , I could have worked geometrically, by using that image of $P$ and point $A$ and $B$ should be collinear. But here vary two things , $A$ and $B$....

Answer 1

Ah, but we can work geometrically:

Simply reflect $P$ across the two lines.

Can you see what to do next?

Answer 2

Another way.

By Minkowski we obtain: $$PA+PB+AB=\sqrt{(x-2)^2+1}+\sqrt{(y-x)^2+y^2}+\sqrt{(1-y)^2+(2-y)^2}\geq$$ $$\geq\sqrt{(x-2+y-x+1-y)^2+(1+y+2-y)^2}=\sqrt{10}.$$ The equality occurs for $A\left(\frac{5}{3},0\right)$ and $B\left(\frac{5}{4},\frac{5}{4}\right),$ which says that $\sqrt{10}$ is a minimal value.