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- Feb 14, 2012

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- Thread starter anemone
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- #1

- Feb 14, 2012

- 3,812

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- Feb 14, 2012

- 3,812

$x=a+2b+c,\\y=a+b+2c,\\z=a+b+3c$

It is easy to see that $z-y=c$ and $x-y=b-c$, giving $x-y=b-(z-y)$ or $b=x+z-2y$. Note that $a+3c=2y-x$. By the AM-GM inequality, it follows that

$\dfrac{a+3c}{a+2b+c}+\dfrac{4b}{a+b+2c}+\dfrac{8c}{a+b+3c}\\=\dfrac{2y-x}{x}+\dfrac{4( \dfrac{2y-x}{x} )}{y}- \dfrac{8(z-y)}{z}\\=-17+2\left(\dfrac{y}{x}\right)+4\left(\dfrac{x}{y}\right)+4\left(\dfrac{z}{y}\right)+8\left(\dfrac{y}{z}\right)\\ \ge -17+2\sqrt{8}+2\sqrt{32}\\=-17+12\sqrt{2}$

The equality holds if and only if $\dfrac{2y}{x}=\dfrac{4x}{y}$ and $\dfrac{4z}{y}=\dfrac{8y}{z}$, or $4x^2=2y^2=z^2$. Hence, the equality holds if and only if

$a+b+2c=\sqrt{2}(a+2b+c)\\a+b+3c=2(a+2b+c)$

Solving the above system of equations for $b$ and $c$ in terms of $a$ gives

$b=(1+\sqrt{2})a\\c=(4+3\sqrt{2})a$

We conclude that $\dfrac{a+3c}{a+2b+c}+\dfrac{4b}{a+b+2c}+\dfrac{8c}{a+b+3c}$ has a minimum value of $12\sqrt{2}-17$ if and only if

$(a,\,b,\,c)=(a,\,(1+\sqrt{2})a,\,(4+3\sqrt{2})a)$