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Find the minimum value of $\frac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$ for $x\gt0$.

In some textbook, the problem is usually tackled by calculus. So I started to investigate the problem using triangle inequality only. The answer is so interesting and simple that the minimum value of $S(x)$ is $\sqrt{10}$ when $x=\frac{1+\sqrt{13}}{6}$.

My Question: How many elegant methods are there to find the minimum point?

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We are going to find the minimum value of} $$ S(x)=\dfrac{\sqrt{x^{4}+x^{2}+2 x+1}+\sqrt{x^{4}-2 x^{3}+5 x^{2}-4 x+1}}{x}$$ geometrically using Triangle Inequality.

$$\displaystyle \begin{aligned} S(x)&=\sqrt{x^{2}+1+\frac{2}{x}+\frac{1}{x^{2}}} +\sqrt{x^{2}-2 x+5-\frac{4}{x}+\frac{1}{x^{2}}} \\&=\sqrt{x^{2}+\left(\frac{1}{x}+1\right)^{2}} +\sqrt{(x-1)^{2}+\left(\frac{1}{x}-2\right)^{2}}\end{aligned}$$ $\textrm{which is the sum of distances of any point }P\text{ on a rectangular hyperbola from }(0,-1)$ $\textrm{and }(1,2) \textrm{ as shown below:}$

enter image description here

Since $S(x) =A P+P B \geqslant A B =\sqrt{(1-0)^{2}+(2-(-1))^{2}}=\sqrt{10},$ therefore the minimum value of S(x) is $\sqrt{10},\\ $ when and only when A, B and P are collinear$ \quad \textrm{ i.e. } \dfrac{2-\frac{1}{x}}{1-x}=\dfrac{2-(-1)}{1-0 } \Leftrightarrow \dfrac{2 x-1}{1-x}=3 x \Leftrightarrow 3 x^{2}-x-1=0 \Leftrightarrow x=\dfrac{1+\sqrt{13}}{6}. $

We can now conclude that the minimum value of S(x) is $\sqrt{10}$ when $x=\dfrac{1+\sqrt{13}}{6}.$

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  • $\begingroup$ Beautiful (+1). A question for you. To generalize, how much is this methodology dependent on the exact coefficients of the polynomial that you have under the square root ? $\endgroup$
    – Thomas
    Feb 8 at 11:33
  • $\begingroup$ Good idea!Thank you. $\endgroup$
    – Lai
    Feb 9 at 13:31

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