Question

Cho a,b,c dương thoả mãn: abc≥1. CMR: 

\(\left(a+\dfrac{1}{a+1}\right).\left(b+\dfrac{1}{b+1}\right).\left(c+\dfrac{1}{c+1}\right)\ge\dfrac{27}{8}\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM ta có:

$\text{VT}=[\frac{a+1}{4}+\frac{1}{a+1}+\frac{3}{4}a-\frac{1}{4}][\frac{b+1}{4}+\frac{1}{b+1}+\frac{3}{4}b-\frac{1}{4}][\frac{c+1}{4}+\frac{1}{c+1}+\frac{3}{4}c-\frac{1}{4}]$

$\geq [2\sqrt{\frac{1}{4}}+\frac{3}{4}a-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}b-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}c-\frac{1}{4}]$
$=\frac{3}{4}(a+1).\frac{3}{4}(b+1).\frac{3}{4}(c+1)$
$=\frac{27}{64}(a+1)(b+1)(c+1)$

$\geq \frac{27}{64}.2\sqrt{a}.2\sqrt{b}.2\sqrt{c}$

$=\frac{27}{64}.8\sqrt{abc}\geq \frac{27}{64}.8=\frac{27}{8}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$