1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $$(a+1)(b+1)(c+1)\geq 8.$$

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.