Prove that $\frac1{a(1+b)}+\frac1{b(1+c)}+\frac1{c(1+a)}\ge\frac3{1+abc}$

Question

I tried doing it with CS-Engel to get $$ \frac{1}{a(1+b)}+\frac{1}{b(1+c)}+\frac{1}{c(1+a)} \geq \frac{9}{a+b+c+ a b+b c+a c} $$ I thought that maybe proof that $$ \frac{1}{a+b+c+a b+b c+a c} \geq \frac{1}{3(1+a b c)} $$ or $$ 3+3 a b c \geq a+b+c+a b+b c+a c $$, but I don't know how

Answer 1

For positive variables by AM-GM we obtain: $$\sum_{cyc}\frac{1}{a(1+b)}=\frac{1}{1+abc}\sum_{cyc}\frac{1+abc}{a(1+b)}=\frac{1}{1+abc}\left(\sum_{cyc}\frac{1+abc}{a(1+b)}+1-1\right)=$$ $$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab+abc}{a(1+b)}-\frac{3}{1+abc}=$$ $$=\frac{1}{1+abc}\sum_{cyc}\frac{1+a+ab(1+c)}{a(1+b)}-\frac{3}{1+abc}=$$ $$=\frac{1}{1+abc}\sum_{cyc}\left(\frac{1+a}{a(1+b)}+\frac{b(1+c)}{1+b}\right)-\frac{3}{1+abc}\geq$$ $$\geq\frac{6}{1+abc}\sqrt[6]{\prod_{cyc}\left(\frac{1+a}{a(1+b)}\cdot\frac{b(1+c)}{1+b}\right)}-\frac{3}{1+abc}=\frac{3}{1+abc}.$$

Answer 2

As in Michael's solution, this uses the same idea of "+1" to split up the fraction.
Written up this way, it might seem like a more natural approach.

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WTS

$$ \sum \frac{ 1 + abc } { a (1+b)} \geq 3 $$

$$\sum \frac{ 1 + abc + a + ab } { a (1+b) } \geq 6$$

$$ \sum \frac{ ab ( 1 + c) + (1+a ) } { a (1+b) } \geq 6$$

$$ \sum \frac{ ab (1+c) } { a(1+b) } + \frac{ (1+b) } { b (1 + c) } \geq 6$$

Applying AM-GM to all 6 terms, the result follow.
From the conditions, we can easily deduce that the equality case is only $ a = b = 1$.