<s>[INST] Find the $x$ value for when the expression is takes the minimum value 
Let $E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}.$ Find the $x$ value for when this expression take it's minimum value.

Using the derivative it's pretty hard... and I think not indicated for this problem... I thought about using the inequality of means. First, rewrite:
$E(x)=\sqrt{x^2-2x+5}+\sqrt{x^2-8x+25}$ as $E(x)=\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}$..
But I don't know how to continue...
 [/INST] Using Minkowski's inequality:
$$\sqrt{(x-1)^2+4}+\sqrt{(x-4)^2+9}=\\
\sqrt{(x-1)^2+2^2}+\sqrt{(4-x)^2+3^2}\ge \\
\sqrt{(x-1+4-x)^2+(2+3)^2}=\sqrt{3^2+5^2}=\sqrt{34}.$$
The equality occurs when $(x-1,2)||(4-x,3)$, that is:
$$\frac{3}{2}(x-1)=4-x \Rightarrow x=\frac{11}{5}.$$
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