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i it did the hard way.
on evaluation and simplification we get,
2 + 9abc > 7ab + 7bc + 7ca.
so if we prove this we can prove the above thing.
we start from this (a+b)(ab)^{2} + (b+c)(bc)^{2} + (c+a)(ca)^{2} > 0
if we proceed now we can get there.
Rewrite the problem by substituting 1 by (a+b+c) as having to prove that
Setting and dividing by each factor on both sides by 2, we can further rewrite as
given x+y+z=1, to prove that which is well known and easy as
Multiplying the above three inequalities concludes the solution
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