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[huyqhx9]

Cho a,b,c là các số dương thỏa mạn abc = 1 .Chứng minh rằng:
 
$\frac{a^{3}}{\left ( 1+a \right )(1+b)} +\frac{b^{3}}{(1+b)(1+c)} + \frac{c^{3}}{(1+c)(1+a)}\geq \frac{3}{4}$

Edited by huyqhx9, 05-07-2016 - 09:46.

[Baoriven]

Ta có: $\frac{a^3}{(1+a)(1+b)}+\frac{1+a}{8}+\frac{1+b}{8}\geq \frac{3}{4}a$

Suy ra: 

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

Dấu bằng xảy ra khi $a=b=c=1$