Find minimum of function $\frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$

Question

Find minimum of function $f(x) = \frac{\left| x-12\right| }{5}+\frac{\sqrt{x^2+25}}{3}$

I tried compute min using by definition of abs function. I consider two cases:

1. when $x > 12$ we have: $f_{1}(x) = \frac{x-12}{5}+\frac{\sqrt{x^2+25}}{3}$ using standard method $f_{1}'(x)=0$ for $x_{0}=-15/4$ but $x_{0}$ is not in $ [12, \infty]$
2. when $x \leq 12$ we have $f_{2}(x) = \frac{-x+12}{5}+\frac{\sqrt{x^2+25}}{3}$ similarly $f_{2}'(x) = 0$, and $x = 15/4$

So using observation from this method I computed minimum of $f(15/4) = 56/15$.

Does it be a correct way?

Answer 1

We can use C-S and the triangle inequality:

$$\frac{|x-12|}{5}+\frac{\sqrt{x^2+25}}{3}=\frac{|x-12|}{5}+\frac{\sqrt{(3^2+4^2)(x^2+5^2)}}{15}\geq$$ $$\geq \frac{|x-12|}{5}+\frac{|3x+20|}{15}=\left|\frac{12}{5}-\frac{x}{5}\right|+\left|\frac{4}{3}+\frac{x}{5}\right|\geq $$ $$\geq \left|\frac{12}{5}-\frac{x}{5}+\frac{4}{3}+\frac{x}{5}\right|=\frac{56}{15}.$$ The equality occurs for $(3,4)||(x,5)$ and $\left(\frac{12}{5}-\frac{x}{5}\right)\left(\frac{4}{3}+\frac{x}{5}\right)\geq0,$ which gives $x=\frac{15}{4},$

which says that we got a minimal value.

Answer 2

Your reasoning is spot on!

Indeed, since both terms are never negative, it is certain that there is a minimum. But you have exhibited the only possibility. It follows that that is indeed the minimum value of the function.