# Minimum value of $\sqrt{x^4 + 3x^2 - 6x + 10} + \sqrt{x^4 - 5x^2 + 9}$ without using calculus?

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.

• 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.... – achille hui Nov 10 '20 at 19:04
• @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. – Dave L. Renfro Nov 10 '20 at 19:32
• @DaveL.Renfro instead of algebra, there is a geometric way to get the minimum. see my answer. – achille hui Nov 10 '20 at 19:41

## 1 Answer

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 $$AB = \sqrt{(0-3)^2 + (3-(-1))^2} = \sqrt{3^2 + 4^2} = 5$$

• (+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. – Dave L. Renfro Nov 10 '20 at 19:47