Problem:
Let a,b, and c be nonnegative real numbers such that
a2+b2+c2+abc=4
Prove that
0≤ab+bc+ca−abc≤2
Solution:
First Solution: From the condition, at least one of a,b, and c does not exceed 1, say a≤1. Then
ab+bc+ca−abc=a(b+c)+bc(1−a)≥0
Now we prove the upper bound. Let us note that at least two of the three numbers a.b. and c are both greater than or equal to 1 or less than or equal to 1 . Without loss of generality. we assume that the numbers with this property are b and c. Then we have
(1−b)(1−c)≥0(1)
The given equality a2+b2+c2+abc=4 and the inequality b2+c2≥2bc imply that
a2+2bc+abc≤4, or bc(2+a)≤4−a2
Dividing both sides of the last inequality by 2+a yields
The last equality holds if and only if b=c and a(1−b)(1−c)=0. Hence equality for the upper bound holds if and only if (a,b,c) is one of the triples (1,1,1).(0,2,2). (2,0,2), and (2,2,0). Equality for the lower bound holds if and only if (a,b,c) is one of the triples (2,0,0),(0,2,0), and (0.0.2).
Second Solution: The proof for the lower bound is the same as in the first solution. Now we prove the upper bound. It is clear that a,b.c≤2. If abc=0, then the result is trivial. Suppose that a,b,c>0. Solving for a yields
a=2−bc+b2c2−4(b2+c2−4)=2−bc+(4−b2)(4−c2)
This ask for the trigonometric substitution b=2sinu and c=2sinv, where 0∘<u⋅v<90∘. Then
a=2(−sinusinv+cosucosv)=2cos(u+v)
and if we set u=B/2 and v=C/2. then a=2sin(A/2),b=2sin(B/2), and c=2sin(C/2), where A,B, and C are the angles of a triangle. We have
ab=4sin2Asin2B=2sinAtan2AsinBtan2B=2sinAtan2BsinBtan2A≤sinAtan2B+sinBtan2A (by the AM-GM inequality) =sinAcot2A+C+sinBcot2B+C