What is the minimum value of a radical sum? How would you find the minimum value of $\sqrt {x^2+16} + \sqrt {x^2-12x+37}$ through algebraic manipulation? Graphing will clearly show that the minimum value is 4.8, but how do you get the answer manually?
In general, how would you find the minimum value(if it exists) of the expression $\sqrt[n_1] {P_1(x)} + \sqrt[n_2] {P_2(x)} + \sqrt[n_3] {P_3(x)} + \cdots + \sqrt[n_i] {P_i(x)}$, where $P_1(x), P_2(x), P_3(x), \ldots, P_i(x)$ are polynomials, and $n_1, n_2, n_3, \ldots,n_i$ are whole numbers? On the other hand, how would you prove that no minimum value for such a sum exists?
 A: Solution to the first minimization problem:
$\sqrt{x^2+16}$ is the distance from the point $(x, 0)$ to $(0, 4)$
and
$\sqrt{x^2-12x+37}=\sqrt{(x-6)^2+(1)^2}$ is the distance from the point$(x, 0)$ to $(6, -1)$
Thus the sum is the distance from $(0, 4)$ to $(x, 0)$ to $(6, -1)$ which by the triangle inequality is at minimum the distance from $(0, 4)$ to $(6, -1)$, which occurs at the intersection of the line through the points $(0, 4)$ and $(6, -1)$ with the $x$-axis.  The line through those two points is, using slope-intercept form,
$$y=mx+4$$
To find $m$, we know that $(6, -1)$ is on the line, so 
$$-1=6m+4$$
$$m=-\frac{5}{6}$$
so the equation of the line is
$$y=-\frac{5}{6}x+4$$
which intercepts the $x$-axis when $y=0$, which is when
$$0=-\frac{5}{6}x+4$$
$$x=24/5=4.8,$$
as expected.
To do this algebraically, the only thing you have to do is to state the triangle inequality in 2 dimensions using algebra.
I have not done this for more points, but you can do this with the sum of any 2 square roots as long as the coefficients of the $x^2$ terms are positive.
So the strategy is: complete the square within each square root, and use the distance formula to find 2 points on opposite sides of the $x$-axis such that the expression is the sum of the distances from each of the points to $(x, 0)$ and then use the triangle inequality.
A: Another way.
By C-S $$\sqrt{x^2+16}+\sqrt{x^2-12x+37}=$$
$$=\frac{1}{\sqrt{2.44}}\sqrt{(1.2^2+1)^2(x^2+4^2)}+\frac{1}{\sqrt{2.44}}\sqrt{(1.2^2+1^2)((6-x)^2+1^2)}\geq$$
$$=\frac{1}{\sqrt{2.44}}\left(1.2x+4+1.2(6-x)+1\right)=\frac{12.2}{\sqrt{2.44}}=\sqrt{61}.$$
The equality occurs for $(x,4)||(6-x,1)$ or for $x=4.8,$ which says that we got a minimal value.
