<s>[INST] Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ Prove that $\sqrt{(x-2)^2+y^2}+\sqrt{(x-4)^2+(y-2)^2}+\sqrt{(x-2)^2+(y-5)^2}+\sqrt{(x-1)^2+(y-2)^2}\geq8$ for $x,y\in\mathbb{R}$. I used this Proof of an inequality about sequences to solve it and it worked, but I have no ideea how to interpret it geometrically. Any help, please? Thanks in advance.
 [/INST] Let's $A(2,0), B(4,2),C(2,5), D(1,2)$ and $G(x,y)$ on the plane $Oxy$. The left hand side of the inequality is the sum  $$\mathbf{S} =GA+GB+GC+GD$$
The inequality becomes: find $G$ such that the sum $\mathbf{S}$ is minimized.
We have
$$GA+GC \ge AC$$ $$GB+GD \ge BD$$
then $$S = GA+GB+GC+GD \ge AC+BD = \sqrt{2^2+5^2}+3 = \sqrt{29}+3$$
The sum is then minimized if $G$ is at $I$, the intersection of two lines: $AC$ and $BD$.
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