Let $a,b$ and $c$ be the positive real numbers. Determine the minimum value of $$\frac{a+3c}{a+2b+c} + \frac{4b}{a+b+2c} - \frac{8c}{a+b+3c}.$$
2026-03-29 11:02:13.1774782133
Determine the minimum value of the expression, for $a,b$ , $c$ $\in\ \textbf{R}^+$
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Let $a+2b+c=x$, $a+b+2c=y$ and $a+b+3c=z$.
Hence, $a=5y-x-3z$, $b=x-2y+z$, $c=z-y$ and by AM-GM $$\frac{a+3c}{a+2b+c} + \frac{4b}{a+b+2c} - \frac{8c}{a+b+3c}=4\left(\frac{z}{y}+\frac{2y}{z}\right)+2\left(\frac{2x}{y}+\frac{y}{x}\right)-17\geq $$ $$\geq8\sqrt2+4\sqrt2-17=12\sqrt2-17.$$ The equality occurs for $\frac{z}{y}=\frac{2y}{z}$ and $\frac{2x}{y}=\frac{y}{x}$, which says that the equality indeed occurs.
Id est, the answer is $12\sqrt2-17$.