Subtracting 2(ab+bc+ca) from both sides, this gives
2(ab+bc+ca)+4c2≥a2+b2+c2
as desired.
Remark. The equality in the AM-GM step occurs if and only if c(a+b+c)=ab. Solving for a+b+c and substituting into the condition a+b+c=43abc, this implies 8c2=ab. Substituting this back into the equation c(a+b+c)=ab, we conclude that
c(a+b+c)=8c2⟹a+b=7c
We then have
a−b=±(a+b)2−4ab=±49c2−32c2=±17c
It follows that {2a,2b}={(7−17)c,(7+17)c}. Hence, equality holds if and only if (a,b,c) is a permutation of
((7−17)r,(7+17)r,2r)
for some positive real number r.
Second solution. Suppose, as above, that c=min(a,b,c), and write A=a/c,B=b/c, and D=A+B. The given condition becomes A+B+1=43AB, or equivalently, AB=(D+1)3/64. In terms of A and B, the problem asks us to prove that
2(AB+A+B)+4≥A2+B2+1
which can be rearranged as
2(A+B)+3−(A+B)2+4AB≥0
After substituting in D, this inequality becomes
2D+3−D2+(D+1)3/16≥0
Since the left-hand side factors as (D+1)(D−7)2/16, the inequality always holds.
Third solution: Assuming that c=min(a,b,c) and by adding 2(ab+bc+ca) to both sides, our inequality becomes
4c(a+b+c)+4ab≥(a+b+c)2
Since both the given condition and the desired claim are homogeneous, we may assume without loss of generality that a+b+c=8, so our task is to prove that if ab=8/c, then 32c+4ab≥64. This clearly holds, since for any positive real number c we have 32(c+c1)≥64.