$f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$. Let $f(x,y) = \sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}-\sqrt{x^2+y^2}-\sqrt{(x-1)^2+y^2}\;\;,x,y\in \mathbb{R}$.
Then Max. of $f(x,y)$.
$\underline{\bf{My\;Try}}::$ We can convert into Complex no. form ...
Let $z=x+iy$, Then $f(z) = \left|z-i\right|+\left|z-3-4i\right|-\left|z\right|-\left|z-1\right|$
Now for Max., we use $\triangle$ Inequality,
But I did not understand in which  pair i have used $\triangle$ inequality.
Help Required
Thanks

let $A(0,1),B(3,4),C(0,0),D(1,0),P(x,y)$, we want $PA+PB-PC-PD$ have max ,but $PA-PC\le AC$, the "=" will hold when $P$ is on $AC$, $PB-PD \le BD$,the "=" will hold when $P$ is $BD$, so MAX is $AC+BD=1+\sqrt{(3-1)^2+(4-0)^2}=1+2\sqrt{5}$,when $P(0,-2)$
