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

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!

Let $$AB=1$$, $$A(1,0),$$ $$B(0,0)$$, $$C(0,1)$$ and $$P(x,y)$$.
Thus, by Minkowski: $$PA+PB+\sqrt2PC=\sqrt{(x-1)^2+y^2}+\sqrt{x^2+y^2}+\sqrt{2(x^2+(y-1)^2)}=$$ $$=\sqrt{(x-1)^2+(-y)^2}+\sqrt{(-y)^2+(-x)^2}+\sqrt{(-x+y-1)^2+(x+y-1)^2}\geq$$ $$\geq\sqrt{(x-1-y-x+y-1)^2+(-y-x+x+y-1)^2}=\sqrt5=\sqrt5BC.$$
• @user628755 I got that the equality occurs on the circle $\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{36}.$ Commented Jun 26, 2019 at 5:38