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Hi mathematics stack exchange, what is the minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$? I know how to solve this problem using calculus, you take a derivative, but I am wondering if there is an elementary method to find the minimum using precalculus methods.

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    $\begingroup$ Interesting. With help of a CAS, the minimum value is the integer $5$ at an irrational $x = \frac{\sqrt{31}-2}{3}$. there should be a trick to get this.... $\endgroup$ – achille hui Nov 10 '20 at 19:04
  • $\begingroup$ @achille hui: I notice that $x^4 + 3x^2 - 6x + 10 = (x^2 - 2x + 2)(x^2 + 2x + 5)$ (trying a factorization of the form $(x^2 + ax + 2)(x^2 + bx + 5)$ leads to equations that can be solved for $a$ and $b)$ and $x^4 - 5x^2 + 9 = x^4 + 6x^2 + 9 - 6x^2 = (x^2 + 3)^2 - 6x^2,$ which as a difference of squares can be factored as $(x^2 + 3 + x\sqrt{6})(x^2 + 3 - x\sqrt{6}),$ but I don't see how to make use of this. $\endgroup$ – Dave L. Renfro Nov 10 '20 at 19:32
  • $\begingroup$ @DaveL.Renfro instead of algebra, there is a geometric way to get the minimum. see my answer. $\endgroup$ – achille hui Nov 10 '20 at 19:41
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Let $y = x^2$, notice

$$\begin{align} x^2 + (y-3)^2 & = x^2 + (x^2-3)^2 = x^4 - 5x^2+9\\ (x-3)^2+(y+1)^2 & = (x-3)^2 + (x^2 + 1)^2 = x^4 - 3x^2 -6x + 10\end{align}$$

The problem at hand can be rephrased as:

Given $A = (0,3)$, $B = (3,-1)$ and $P = (x,y)$ be a point on the parabola $y = x^2$. What is the minimum value of $AP + PB$?

If one make a plot of the parabola $y = x^2$, one will notice the parabola intersect with the line segment $AB$, this means the minimum value of $AP + PB$ is $$AB = \sqrt{(0-3)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = 5$$

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    $\begingroup$ (+1) Totally awesome! I would never have thought of this approach, but it strikes me as something that might be semi-standard for certain math contest-type problems. If so, it's not (yet, at least) in my math toolbox, but I might try playing around with this some to make it so. $\endgroup$ – Dave L. Renfro Nov 10 '20 at 19:47

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