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Let $a_1$, $a_2$, $a_3$ be three different real numbers. Define real numbers $b_1$ $b_2$ $b_3$ as $$b_1 = (1+ \frac{a_1a_2}{a_1-a_2})(1+ \frac{a_1a_3}{a_1-a_3}) $$ $$b_2 = (1+ \frac{a_2a_1}{a_2-a_1})(1+ \frac{a_2a_3}{a_2-a_3}) $$ $$b_3 = (1+ \frac{a_3a_1}{a_3-a_1})(1+ \frac{a_3a_2}{a_3-a_2}) $$ Prove that $$ 1+ |a_1b_1+a_2b_2+a_3b_3|\le (1+|a_1|)(1+|a_2|)(1+|a_3|)$$

I have no idea how to simplify the left side.. but the right side could perhaps be shrunk to $8\sqrt{|a_1a_2a_3|}$?

This question is classified as "variable substitution".. could this help?

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By setting $g_i=\frac{1}{a_i}$ we have: $$ b_1 = \left(1+\frac{1}{g_2-g_1}\right)\left(1+\frac{1}{g_3-g_1}\right)=1+\frac{g_3+g_2-2g_1+1}{(g_2-g_1)(g_3-g_1)}$$ $$ b_2 = \left(1+\frac{1}{g_1-g_2}\right)\left(1+\frac{1}{g_3-g_2}\right)=1+\frac{g_1+g_3-2g_2+1}{(g_1-g_2)(g_3-g_2)}$$ $$ b_3 = \left(1+\frac{1}{g_1-g_3}\right)\left(1+\frac{1}{g_2-g_3}\right)=1+\frac{g_1+g_2-2g_3+1}{(g_1-g_3)(g_2-g_3)}$$ and we have to prove that: $$ \left|\frac{b_1}{g_1}+\frac{b_2}{g_2}+\frac{b_3}{g_3}\right|\leq \frac{1}{|g_1|}+\frac{1}{|g_2|}+\frac{1}{|g_3|}+\frac{1}{|g_1 g_2|}+\frac{1}{|g_1 g_3|}+\frac{1}{|g_2 g_3|}+\frac{1}{|g_1 g_2 g_3|}\tag{1}$$ or: $$\left|b_1 g_2 g_3+ b_2 g_1 g_3 + b_3 g_1 g_2\right|\leq |g_2 g_3|+|g_1 g_3|+|g_1 g_2|+|g_1|+|g_2|+|g_3|+1\tag{2}$$ hence it is sufficient to prove that: $$\left|g_2 g_3\frac{g_3+g_2-2g_1+1}{(g_2-g_1)(g_3-g_1)}+g_1 g_3\frac{g_1+g_3-2g_2+1}{(g_1-g_2)(g_3-g_2)}+g_1 g_2 \frac{g_1+g_2-2g_3+1}{(g_1-g_3)(g_2-g_3)}\right|\leq |g_1|+|g_2|+|g_3|+1 \tag{3}$$ or: $$\left|(g_1+g_2+g_3+1)\sum_{cyc}\frac{g_2 g_3}{(g_2-g_1)(g_3-g_1)}-3g_1g_2g_3\sum_{cyc}\frac{1}{(g_2-g_1)(g_3-g_1)}\right|\leq |g_1|+|g_2|+|g_3|+1$$ or: $$\left|(g_1+g_2+g_3+1)\right|\leq |g_1|+|g_2|+|g_3|+1\tag{4}$$ that is quite trivial.

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