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$a$, $b$, $c$ are three positives and $m$, $n$, $p$, $x$, $y$, $z$ are positive parameters . Find the minimum value of $abc$ such that the following inequation is correct. $$\large \dfrac{x}{x + ma} + \dfrac{y}{y + nb} + \dfrac{z}{z + pc} \le 1$$

The problem asked for the maximum value of $abc$, not the minimum value. So it confused me so much when I first solved the problem.

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Let $\frac{ma}{x}=u,$ $\frac{nb}{y}=v$ and $\frac{pc}{z}=w$.

Thus, $$1\geq\large \dfrac{x}{x + ma} + \dfrac{y}{y + nb} + \dfrac{z}{z + pc}=\sum_{cyc}\frac{1}{1+u}$$ or $$\prod_{cyc}(1+u)\geq\sum_{cyc}(1+u)(1+v),$$ which by AM-GM gives $$uvw\geq2+\sum_{cyc}u\geq2+3\sqrt[3]{uvw},$$ which gives $$(\sqrt[3]{uvw}-2)(\sqrt[3]{uvw}+1)^2\geq0$$ or $$uvw\geq8$$ or $$\frac{mnpabc}{xyz}\geq8$$ or $$abc\geq\frac{8xyz}{mnp}.$$ The equality occurs for $u=v=w=2,$ which says that $\frac{8xyz}{mnp}$ is a minimal value of $abc$.

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The answer is brought to you by my math teacher.

By the AM-GM inequality, we have that

$$\left \{ \begin{aligned} 2\sqrt{\dfrac{xy}{(x + ma)(y + nb)}} &\le \dfrac{x}{x + ma} + \dfrac{y}{y + nb} = \dfrac{pc}{z + pc}\\ 2\sqrt{\dfrac{yz}{(y + nb)(z + pc)}} &\le \dfrac{y}{y + nb} + \dfrac{z}{z + pc} = \dfrac{ma}{x + ma}\\ 2\sqrt{\dfrac{zx}{(z + pc)(x + ma)}} &\le \dfrac{z}{z + pc} + \dfrac{x}{x + ma} = \dfrac{nb}{y + nb} \end{aligned} \right.$$

$$\implies 8 \cdot \dfrac{xyz}{(x + ma)(y + nb)(z + pc)} \le \dfrac{mnp \cdot abc}{(x + ma)(y + nb)(z + pc)} \iff abc \ge 8 \cdot \dfrac{xyz}{mnp}$$

The equal sign occurs when

$$\dfrac{x}{x + ma} = \dfrac{y}{y + nb} = \dfrac{z}{z + pc} = \dfrac{1}{3} \iff \dfrac{x + ma}{x} = \dfrac{y + nb}{y} = \dfrac{z + pc}{z} = 3$$

$$\iff \dfrac{ma}{x} = \dfrac{nb}{y} = \dfrac{pc}{z} = 2 \iff \left \{ \begin{aligned} a &= \dfrac{2x}{m}\\ b &= \dfrac{2y}{n}\\ c &= \dfrac{2z}{p} \end{aligned} \right.$$

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