GTNN

Ta có:

$Q=\frac{2ab+a+b+c(ab-1)}{(a+1)(b+1)(c+1)}$

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

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

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

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

Từ $\ \ c \geq b+1$ và $\ \ a+b \geq c$ suy ra $\ \ a+b \geq b+1$  kéo theo $\ \ b\geq a\geq 1$

Nếu:  $b \geq 2$

Do  $\ \ a+b \geq c$  nên $Q \geq \frac{1}{1+a+b}+\frac{b}{1+b}-\frac{1}{1+a}=\frac{a}{1+a}+\frac{b}{1+b}-\frac{a+b}{1+a+b}$

Vì $\ \ b \geq 2$ ta có $\ \ \frac{b}{1+b}-\frac{a+b}{1+a+b} \geq \frac{2}{3}-\frac{a+2}{a+3}$

Suy ra $Q \geq \frac{2}{3}+\frac{a}{1+a}-\frac{a+2}{a+3}=\frac{(a-1)(a+5)}{4(a+1)(a+3)}+\frac{5}{12} \geq \frac{5}{12}$

Nếu:  $b \leq 2$

Do $a+b \geq c$ kéo theo $\ \ Q \geq \frac{1}{1+c}+\frac{b}{1+b}-\frac{1}{1+c-b}$

Vì $c \geq 3$  nên  $\frac{1}{1+c}-\frac{1}{1+c-b} \geq \frac{1}{4}-\frac{1}{4-b}$

Suy ra $Q \geq \frac{1}{4}+\frac{b}{1+b}-\frac{1}{4-b}=\frac{5}{6} \frac{(2-b)(b-1)}{(b+1)(4-b)}+\frac{5}{12} \geq \frac{5}{12}$

Vậy GTNN của $Q=\frac{5}{12}$ .Đẳng thức xảy ra khi và chỉ khi $a=1 , b=2 , c=3$