# prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$

Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that

$$abc \ge 8$$

The book does not provide full solutions but only hints, and the one for this question is that it is similar to this problem:

Let $a,b,c$ be positive numbers with $a+b+c=1$, prove that

$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1) \ge 8$$

again there's no solution provided but what I came up with is the following (you can correct me if it is wrong):

$$a+b=1-c$$ $$a+c=1-b$$ $$b+c=1-a$$

$$(\frac{1}{a}-1)(\frac{1}{b}-1)(\frac{1}{c}-1)=(\frac{1-a}{a})(\frac{1-b}{b})(\frac{1-c}{c})=(\frac{b+c}{a})(\frac{a+c}{b})(\frac{a+b}{c})=(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c})$$

By AM-GM we have for each term:

$$(\frac{b}{a}+\frac{c}{a}) \ge 2\sqrt{\frac{bc}{a^2}}$$ $$(\frac{a}{b}+\frac{c}{b}) \ge 2\sqrt{\frac{ac}{b^2}}$$ $$(\frac{a}{c}+\frac{b}{c}) \ge 2\sqrt{\frac{ab}{c^2}}$$

By multiplying the three inequalities we have

$$(\frac{b}{a}+\frac{c}{a})(\frac{a}{b}+\frac{c}{b})(\frac{a}{c}+\frac{b}{c}) \ge 8 \sqrt{\frac{a^2b^2c^2}{a^2b^2c^2}}=8$$

as desired.

Can someone please provide me a proof for the first problem as I cannot find any way around it, Thanks.

Define

\begin{align*} x&=\frac{1}{1+a}\\ y&=\frac{1}{1+b}\\ z&=\frac{1}{1+c} \end{align*}

Then your problem transforms into: given $x+y+z=1$ prove $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1) \geq 8$. I believe you know the rest.

• Can you please expand – Basem Fouda Aug 28 '16 at 17:14

If the problem states that a, b, and c are positive integers, this is pretty easy. Find a common denominator on the LHS, multiply both sides by the denominator and simplify. Then you get: $$3+2(a+b+c)+ab+ac+bc=1+(a+b+c)+ab+ac+bc+abc$$ Simplify this and you get $$2+a+b+c=abc$$ The lowest possible non-zero integers are 1, 2, and 3, therefore the lowest value abc can have is 8.

If, however, a, b, and c are the lowest possible real numbers, this is a very different problem.

From the condition by AM-GM we obtain: $$\prod\limits_{cyc}\frac{a}{1+a}=\prod\limits_{cyc}\left(\frac{1}{1+b}+\frac{1}{1+c}\right)\geq\prod\limits_{cyc}\frac{2}{\sqrt{(1+b)(1+c)}}=8\prod\limits_{cyc}\frac{1}{1+a}$$ Id est, $abc\geq8$.