Point $X$ lies in a right-angled isosceles $\triangle ABC$ ($\angle ABC = 90^\circ$). Prove that
$AX+BX+\sqrt{2}CX \geq \sqrt{5}AB$
and find for which points $X$ the equality is met.
Use rotation with center $C$ and angle $+90^\circ$
Let $p(C, +90^\circ)$ be a rotation which sends $A \rightarrow A'$, $B \rightarrow B'$, and $X \rightarrow X'$.
Thus we have the following: $\overrightarrow{BX}+\overrightarrow{XX'}+\overrightarrow{X'A'} = \overrightarrow{BA'}$.
Hence $BX+XX'+X'A' \geq BA'$ and since $\angle XCX'=90^\circ$, we have that $XX' = \sqrt{2}CX$.
Combining this with the fact that $A'X'=AX$ we get:
$AX+BX+\sqrt{2}CX \geq BA'=\sqrt{BA^2+AA'^2}=\sqrt{4+1}BA=\sqrt{5}BA$ with which the statement is proved.
The equality is met for points $X$ for which $BX=A'X'$.