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Find minimal value of $f(x) = \sqrt{5x^2 - 40x + 85} + \sqrt{5x^2 - 24x + 53}$.

I can solve it using derivatives. Is there any other way to solve it? For example using some popular inequalities?

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HINT :

We can write $$\frac{f(x)}{\sqrt 5}=\sqrt{(x-4)^2+(0-1)^2}+\sqrt{\left(x-\frac{12}{5}\right)^2+\left(0-\frac{11}{5}\right)^2}.$$

This represents the sum of the distance between $(x,0)$ and $(4,1)$ and the distance between $(x,0)$ and $\left(\frac{12}{5},\frac{11}{5}\right)$.

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  • $\begingroup$ So, $\sqrt{(x-4)^2+(0-1)^2}+\sqrt{\left(x-\frac{12}{5}\right)^2+\left(0-\frac{11}{5}\right)^2}$ from inequality of triangle is higher than 2. What's next? $\endgroup$ – user128409235 Nov 5 '15 at 19:04
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    $\begingroup$ @user128409235: The sum is minimum when $(x,0)$ is on the line passing through $(4,1)$ and $(12/5,\color{red}{-}11/5)$. (Why?) $\endgroup$ – mathlove Nov 5 '15 at 19:08
  • $\begingroup$ I understand. Thank you :). $\endgroup$ – user128409235 Nov 5 '15 at 19:19
  • $\begingroup$ @user128409235: You are welcome. $\endgroup$ – mathlove Nov 5 '15 at 19:20
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As suggested by mathlove, if we set $A=\left(\frac{12}{5},\frac{11}{5}\right), B=(x,0)$ and $C=\left(4,1\right)$, we have: $$ f(x) = \sqrt{5}\left( AB+BC \right) $$ that is minimized when $B$ belongs to the $AC$-line.

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