Chứng minh Bất đẳng thức sau:
$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}\leq \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$
Với $a,b,c,d\geq 0$.
Edited by datmc07061999, 28-08-2014 - 11:17.
$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}\leq \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$
Started By datmc07061999, 27-08-2014 - 20:46
1 user(s) are reading this topic
0 members, 1 guests, 0 anonymous users
