$\sqrt{5}BA \leq PA +PB+\sqrt{2}PC$

Question

Let $ABC$ be an isosceles right triangle ($\angle B=90^o$) and a point $P$ in its plane. Prove the inequality $\sqrt{5}BA \leq PA +PB+\sqrt{2}PC$. Find all poins $X$ for which the equality holds.

It might be the most difficult problem I have ever met. First, I thought it has a relation to the lemma $\frac{PA}{a}+\frac{PB}{b}+\frac{PC}{c}\geq \sqrt 3$ . But no! The equality holds differently from the problem. I try to create a new isosceles right triangle with side AC but it does not work. Any idea for this, thank!

Answer 1

Let $AB=1$, $A(1,0),$ $B(0,0)$, $C(0,1)$ and $P(x,y)$.

Thus, by Minkowski: $$PA+PB+\sqrt2PC=\sqrt{(x-1)^2+y^2}+\sqrt{x^2+y^2}+\sqrt{2(x^2+(y-1)^2)}=$$ $$=\sqrt{(x-1)^2+(-y)^2}+\sqrt{(-y)^2+(-x)^2}+\sqrt{(-x+y-1)^2+(x+y-1)^2}\geq$$ $$\geq\sqrt{(x-1-y-x+y-1)^2+(-y-x+x+y-1)^2}=\sqrt5=\sqrt5BC.$$