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Question

Let a,b,c be positive real numbers such that a3+b3=c3. Prove that a2+b2c2>6(ca)(cb).

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Solution

The given inequality may be written in the form
7c26(a+b)c(a2+b26ab)<0.
Putting x=7c2,y=6(a+b)c,z=(a2+b26ab), we have to prove that x+y+z<0.
Observe that x,y,z are not all equal (x>0,y<0).
Using the identity x3+y3+z33xyz=12(x+y+z)[(xy)2+(yz)2+(zx)2],
we infer that it is sufficient to prove x3+y3+z33xyz<0.
Substituting the values of x,y,z, we see that this is equivalent to
343c6216(a+b)3c3(a2+b26ab)3126c3(a+b)(a2+b26ab)<0.
Using c3=a3+b3, this reduces to 343(a3+b3)2216(a+b)3(a3+b3)(a2+b26ab)3126((a3+b3)(a+b)(a2+b26ab)<0.
This may be simplified (after some tedious calculations) to,
a2b2(129a2254ab+129b2)<0.
But 129a2254ab+129b2=129(ab)2+4ab>0. Hence the result follows.
Remark: The best constant θ in the inequality a2+b2c2θ(ca)(cb), where a,b,c
are positive reals such that a3+b3=c3, is θ=2(1+21/3+21/3).

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