$A=\dfrac{2ab+a+b+c(ab-1)}{(a+1)(b+1)(c+1)}$

Bài toán:Cho $a,b,c >0$ thỏa mãn: $a\leq b\leq 3\leq c$ ; $c\geq b+1$ ;$a+b\geq c$

Tìm GTNN của biểu thức:

$A=\dfrac{2ab+a+b+c(ab-1)}{(a+1)(b+1)(c+1)}$

Lời giải:

Có:

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

Ta sẽ chứng minh:
$Q\geq \dfrac{1}{1+1}+\dfrac{2}{1+2}-\dfrac{3}{1+3}=\dfrac{5}{12}$

$\Leftrightarrow \left (\dfrac{3}{1+3}-\dfrac{c}{1+c}  \right )+\left ( \dfrac{b}{b+1}-\dfrac{2}{1+2} \right )+\left ( \dfrac{a}{1+a}-\dfrac{1}{1+1} \right )\geq 0$

$\Leftrightarrow \dfrac{3-c}{4(c+1)}+\dfrac{b-2}{3(b+1)}+\dfrac{a-1}{2(1+a)}\geq 0$

$\Leftrightarrow (3-c)\left [ \dfrac{1}{4(c+1)}-\dfrac{1}{3(b+1)} \right ]+\left [ (3-c)+(b-2) \right ]\left [ \dfrac{1}{3(b+1)}-\dfrac{1}{2(1+a)} \right ]+$
$[(3-c)+(b-2)+(a-1)]\dfrac{1}{2(1+a)}\geq 0$

$\Leftrightarrow (3-c).\dfrac{3b-4c-1}{12(b+1)(c+1)}+(b+1-c).\dfrac{2a-3b-1}{6(b+1)(a+1)}+$
$(a+b-c).\dfrac{1}{2(a+1)}=0~~(*)$

Vì $\left\{\begin{matrix}c\geq 3;~~ 0<b\leq c  &  & \\ b+1\leq c;~~ 0<a\leq b  &  & \\ a+b\geq c;~~0<a \end{matrix}\right.$

nên $(*)$ luôn đúng.

Vậy $Q\geq \dfrac{5}{12}$ khi $a=1;b=2;c=3$