MHB Find the sum of all possible values

[anemone]

$a,\,b$ and $c$ are positive real numbers such that $25bc+9ac+ab=9abc$ and $a+b+c=9$.

Find the sum of all possible values of $abc$.

 

[Opalg]

$a,\,b$ and $c$ are positive real numbers such that $25bc+9ac+ab=9abc$ and $a+b+c=9$.

Find the sum of all possible values of $abc$.

[sp]Let $a=5x$, $b=3y$ and $c=z$. Then (after dividing the first one by $abc$) the equations become $$\frac5x + \frac3y + \frac1z = 9, \qquad 5x + 3y + z = 9.$$ Add those, to get $5\Bigl(x+\dfrac1x\bigr) + 3\Bigl(y+\dfrac1y\bigr) + \Bigl(z+\dfrac1z\bigr) = 18.$

For $t>0$ the minimum value of $t + \dfrac1t$ is 2, attained only when $t=1$. Therefore the minimum value of $5\Bigl(x+\dfrac1x\bigr) + 3\Bigl(y+\dfrac1y\bigr) + \Bigl(z+\dfrac1z\bigr)$ is $2(5+3+1) = 18$, and this is attained only when $x=y=z=1.$ Thus the original equations have the unique solution $a=5$, $b=3$, $c=1$, with $abc = 15$.[/sp]

 

[anemone]

[sp]Let $a=5x$, $b=3y$ and $c=z$. Then (after dividing the first one by $abc$) the equations become $$\frac5x + \frac3y + \frac1z = 9, \qquad 5x + 3y + z = 9.$$ Add those, to get $5\Bigl(x+\dfrac1x\bigr) + 3\Bigl(y+\dfrac1y\bigr) + \Bigl(z+\dfrac1z\bigr) = 18.$

For $t>0$ the minimum value of $t + \dfrac1t$ is 2, attained only when $t=1$. Therefore the minimum value of $5\Bigl(x+\dfrac1x\bigr) + 3\Bigl(y+\dfrac1y\bigr) + \Bigl(z+\dfrac1z\bigr)$ is $2(5+3+1) = 18$, and this is attained only when $x=y=z=1.$ Thus the original equations have the unique solution $a=5$, $b=3$, $c=1$, with $abc = 15$.[/sp]

Thanks Opalg for participating!

Solution of other:

Spoiler

Since $a,\,b,\,c>0$, we can divide the first equation by $abc$ and get

$\dfrac{25}{a}+\dfrac{9}{b}+\dfrac{1}{c}=9$, i.e.

Note that we can borrow and use Cauchy-Schwarz Inequality in the following manner:

$\left(\left(\dfrac{5}{\sqrt{a}}\cdot \sqrt{a}\right)+\left(\dfrac{3}{\sqrt{b}}\cdot \sqrt{b}\right)+\left(\dfrac{1}{\sqrt{c}}\cdot \sqrt{c}\right)\right)^2\le \left(\dfrac{5^2}{a}+\dfrac{3^2}{b}+\dfrac{1^2}{c}\right)(a+b+c)$

$(5+3+1)^2\le \left(\dfrac{5^2}{a}+\dfrac{3^2}{b}+\dfrac{1^2}{c}\right)(9)$

$9\le \dfrac{25}{a}+\dfrac{9}{b}+\dfrac{1}{c}$

But we are told $\dfrac{25}{a}+\dfrac{9}{b}+\dfrac{1}{c}=9$.

Therefore, equality for the Cauchy-Schwarz Inequality above holds when

$\dfrac{5^2}{a^2}=\dfrac{3^2}{b^2}=\dfrac{1^2}{c^2}$ from which we get the unique solution where

$a=5,\,b=3,\,c=1$

Hence the sum of all possible values of $abc$ is $5(3)(1)=15$.