Show that for positive reals a,b,c,
a2b+c+b2c+a+c2a+b≥3a3+3b3+3c32a2+2b2+2c2
What I did was WLOG a+b+c=1 (since the inequality is homogenous)
Then I substituted into the LHS to get ∑cyca21−a≥3a3+3b3+3c32a2+2b2+2c2. Now I'm not sure if I should clear the denominators and expand and try to use Muirhead+Schur? (Clearing the denominators seems quite tedious)?
Any ideas are appreciated.
Answer
your inequalitiy is equivalent to
(a3−2abc+b3+c3)(2a3−a2b−a2c−ab2−ac2+2b3−b2c−bc2+2c3)≥0
since a3+b3+c3≥3abc>2abc
is the first factor positive,
and
2(a3+b3+c3)≥ab(a+b)+ac(a+c)+bc(b+c)
since a3+b3=(a+b)(a2+b2−ab)≥ab(a+b) etc is the second factor also positive.
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