L
lp_qt


Chứng minh rằng với mọi $a;b;c$ dương thì
$\dfrac{1}{a(a+b)}+\dfrac{1}{b(b+c)}+\dfrac{1}{c(c+a)} \ge \dfrac{27}{2(a+b+c)^2}$
$\dfrac{1}{a(a+b)}+\dfrac{1}{b(b+c)}+\dfrac{1}{c(c+a)} \ge \dfrac{27}{2(a+b+c)^2}$