consider (b+c)a and (a+c)b
on applying A.M G.M,
(b+c)a+ (a+c)b ≥ 2√(b+c)(a+c)ab
from this we can say,
(a+b)c + (b+c)a+ (a+c)b ≥ √(b+c)(a+c)ab ...( equality will not hold)
=> ab(a+b) + bc(b+c) + ac(a+c) ≥ ab√cab(b+c)(a+c)
...
=
Try this, it's really appealing....
a,b,c>0
Prove the following inequality:
ab(a+b)+bc(b+c)+ca(c+a)\ge \sum_{cyc}ab\sqrt{\frac{a}{b}\left(b+c \right)\left(c+a \right)}
consider (b+c)a and (a+c)b
on applying A.M G.M,
(b+c)a+ (a+c)b ≥ 2√(b+c)(a+c)ab
from this we can say,
(a+b)c + (b+c)a+ (a+c)b ≥ √(b+c)(a+c)ab ...( equality will not hold)
=> ab(a+b) + bc(b+c) + ac(a+c) ≥ ab√cab(b+c)(a+c)
...
=