How to prove this inequality $\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$ Let $x,y\in R$, show that
$$\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$$

Hint:


*

*$\sqrt{2x^2-6x+5}= \sqrt{(x-1)^2+(x-2)^2}$ is the distance from $A(1,2)$ to $B(x,x)$.

*$\sqrt{2x^2-2xy+y^2}=\sqrt{(x-y)^2+x^2}$ is the distance from $B(x,x)$ to $C(y,0)$.

*$\sqrt{y^2-4y+5}=\sqrt{(y-2)^2+(-1)^2 }$ is the distance from $C(y,0)$ to $D(2,-1)$.


The inequality is then equivalent to
$$AB+BC+CD\ge AD.$$
