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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!

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  • $\begingroup$ Note that $\left(\dfrac {20}3\right)^2=\left(\dfrac{20}3-12\right)^2+16$ $\endgroup$ – J. W. Tanner May 2 at 13:59
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    $\begingroup$ $\sqrt{x^2+a^2}+\sqrt{(c-x)^2+b^2}\ge\sqrt{c^2+(a+b)^2}$; here $a=5, b=4,$ and $c=12$ $\endgroup$ – J. W. Tanner May 2 at 22:34
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$\sqrt{x^2+25}$ is the distance between $(x,0)$ and $(0,5)$, while $\sqrt{(x-12)^2+16}$ is the distance between $(x,0)$ and $(12,-4)$. Hence we need to find $x$ such that the sum of these two distances reaches the minimum. That is exactly the point that the line through $(12,-4)$ and $(0,5)$ meets the $x$-axis. Therefore we get $x = \frac{20}{3}$ and so $y = 12-\frac{20}{3} = \frac{16}{3}$.

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  • $\begingroup$ Why $(0,5)$ and $(12,-4)$? It could be $(0, -5)$ or $(12,4)$ $\endgroup$ – Vasya May 2 at 14:06
  • $\begingroup$ The same. Just find $(x,0)$ such that the sum of two distance reaches the minimum. We will get the same answer if we use these points. $\endgroup$ – Hongyi Huang May 2 at 14:08
  • $\begingroup$ No, it is different if we use $(0, -5)$ and $(12,4)$ because we will never intersect axis $x$. $\endgroup$ – Vasya May 2 at 14:30
  • $\begingroup$ Exactly the same, by symmetry.@Vasya $\endgroup$ – Hongyi Huang May 2 at 14:34
  • $\begingroup$ The question is to find $x$ such that the sum reaches min. So we choose $(0,5)$ and $(12,-4)$ instead of $(0,5)$ and $(12,4)$ because we need the segment to intersect axis $x$, and thus the minimum is got. $\endgroup$ – Hongyi Huang May 2 at 14:37
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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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    $\begingroup$ Equality holds when $\overrightarrow a \parallel \overrightarrow a + \overrightarrow b,$ i.e., $x/12=5/9$, i.e., $x=60/9=20/3$ $\endgroup$ – J. W. Tanner May 3 at 0:32
  • $\begingroup$ Best solution on this page :-) $\endgroup$ – Andreas May 3 at 13:26
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Here's an algebraic solution.

Let $x = z + 20/3$ then $$ \sqrt{x^2+25}+\sqrt{y^2+16}=\sqrt{x^2+25}+\sqrt{(x-12)^2+16}\\ =\sqrt{(z+20/3)^2+25}+\sqrt{(z-16/3)^2+16}\\ = \sqrt{z^2 + 40 z/3 + 625/9}+\sqrt{z^2 - 32 z/3 + 400/9} = f(z) $$ Then $$ f(z)^2= 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \\ = 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81} $$ Since $f(z)$ is always positive, we can claim that $f(z)^2 \ge 15^2 = 225$ so we need to show that $$ 1025/9 + 2 z^2 + 8z/3 + 2 \sqrt{z^2 + 40 z/3 + 625/9}\sqrt{z^2 - 32 z/3 + 400/9} \ge 2025/9\\ \leftrightarrow \sqrt{z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81} \ge 500/9 - z^2 - 4z/3 \\ \leftrightarrow z^4 + (8 z^3)/3 - (85 z^2)/3 - (4000 z)/27 + 250000/81 \ge (500/9 - z^2 - 4z/3)^2 = \\z^4 + (8 z^3)/3 - (328 z^2)/3 - (4000 z)/27 + 250000/81 \\ \leftrightarrow - (85 z^2)/3 \ge - (328 z^2)/3 $$ and this is obviously true, with equality for $z=0$.

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Let

$f = \sqrt{x^2 + 25} + \sqrt{y^2 + 16}$

$g = x+y -12 =0 $

By Lagrange's Undetermined multipliers method,

$F = f + g\lambda = \sqrt{x^2 + 25} + \sqrt{y^2 + 16} + \lambda(x+y -12)$

Differentiating partially w.r.t x and y and using $F_x = 0$ and $F_y = 0 $ at extremum,

$F_x = \frac{x}{\sqrt{x^2 + 25}} +\lambda = 0$

$F_y = \frac{y}{\sqrt{y^2 + 16}} +\lambda = 0$

So, $\frac{x}{\sqrt{x^2 + 25}} = \frac{y}{\sqrt{y^2 + 16}}$

$\frac{x^2}{x^2 + 25} = \frac{y^2}{y^2 + 16}$

$\frac{x^2 + 25}{x^2} = \frac{y^2 + 16}{y^2}$

$1 + \frac{25}{x^2} = 1 + \frac{16}{y^2}$

On solving,

$5y = \pm 4x$

Using this in g,

$x \pm 4x/5 = 12$

$x = 20/3, 60$

Using these values,

$y = 12 - x $

$y = 12 -20/3 = 16/3$ and $y = 12 - 60 = -48$

Using these values of x and y in f we find its value is minimum at $(\frac{20}{3},\frac{16}{3})$.

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    $\begingroup$ I think OP wanted an answer without differentiation $\endgroup$ – J. W. Tanner May 2 at 14:18
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    $\begingroup$ ... and the expression for $F_y$ is missing a $\lambda$. $\endgroup$ – NickD May 2 at 19:23
  • $\begingroup$ Thank you Nick. $\endgroup$ – Ak19 May 3 at 0:44

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