<s>[INST] Minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12$ If $x,y\in\mathbb R^+$, $x+y=12$, what is the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$?
I got the question from a mathematical olympiad competition (of China) for secondary 2 student, so I don't expect an "analysis" answer.
The answer should be $15$, when $x=\frac{20}{3}$ and $ y=\frac{16}{3}$ (just answer, no solution :< ).
I try to use AM-GM inequality, but I couldn't manage to get the answer.
$$\sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}=\sqrt{x^2+25}+\sqrt{x^2-24x+160}$$
Can this help?
I also tried to plot the graph, see here.
Any help would be appreciated. Thx!
 [/INST] We are asked for the minimum value of $\sqrt{x^2+25}+\sqrt{y^2+16}$ if $x+y=12;$
that is, the minimum value of $\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$
Let $\overrightarrow a=(x,5)$ and $\overrightarrow b=(12-x,4)$ in $\mathbb R^2,$ so $\overrightarrow a + \overrightarrow b=(12,9).$
By the triangle inequality, $|\overrightarrow a+\overrightarrow b|\le|\overrightarrow a|+|\overrightarrow b|.$
Therefore, $\mathbf{15}=\sqrt{12^2+9^2}\le\sqrt{x^2+25}+\sqrt{(12-x)^2+16}.$
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