I have read this problem in a book discussing inequalities, and I have a few problems in understanding the proof.

The problem is:

Let $a,b,c$ be positive numbers that satisfy $abc=1$, prove that

$$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)} \ge \frac{3}{4}$$

The solution is:

$$\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)}$$ $$=\frac{(a+1)(b+1)(c+1)-2}{(a+1)(b+1)(c+1)}=1-\frac{2}{(a+1)(b+1)(c+1)} \ge \frac{3}{4}$$

if and only if $(a+1)(b+1)(c+1) \ge 8$ and this last inequality follows immediately from the inequality $(\frac{a+1}{2})(\frac{b+1}{2})(\frac{c+1}{2}) \ge \sqrt{abc} = 1$

I have a problem understanding the last two steps and would like someone to clarify what is actually going on.


The last inequality follows from $a+1\geq 2\sqrt{a}$ (and analogous inequalities for $b$ and $c$), which is true because $(\sqrt{a}-1)^2\geq 0$.

For the step before this, you can rearrange $$1-\frac{2}{(a+1)(b+1)(c+1)} \ge \frac{3}{4}$$ to get $$\frac{1}{4}\geq\frac{2}{(a+1)(b+1)(c+1)}$$ and $$(a+1)(b+1)(c+1)\geq 8.$$

  • $\begingroup$ How do you get $\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)} =\frac{(a+1)(b+1)(c+1)-2}{(a+1)(b+1)(c+1)}$ $\endgroup$ – Basem Fouda Aug 27 '16 at 16:20
  • 1
    $\begingroup$ Make a common denominator $(a+1)(b+1)(c+1)$. The numerator is $a(c+1)+b(a+1)+c(b+1)=ab+ac+bc+a+b+c=(a+1)(b+1)(c+1)-2$, using $abc=1$. $\endgroup$ – pi66 Aug 27 '16 at 17:13

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