Problem 8 · Bridge between symmetric identities and an attainable bound · 5 points
Made by AI and not reviewed by a real person.
Real numbers a, b, and c satisfy a+b+c=0 and a²+b²+c²=6. What is the greatest possible value of a⁶+b⁶+c⁶?
- 72
- 64
- 78
- 54
- 66
Show a hint
- Treat a², b², and c² as three new numbers, and look for what their sum and pairwise products tell you.
- To check whether a bound can actually be reached, set c=t and use (a−b)²≥0 to restrict t.
Show the answer and solution
Answer: E — 66
Since a+b+c=0, we have ab+bc+ca=−3. Also, a²b²+b²c²+c²a²=(ab+bc+ca)²−2abc(a+b+c)=9. The sum-of-cubes identity applied to a², b², c² gives a⁶+b⁶+c⁶=6³−3·6·9+3a²b²c²=54+3a²b²c². Now let c=t. Then a+b=−t and ab=t²−3, so (a−b)²=12−3t²≥0 and t²≤4. Thus 4−a²b²c²=4−t²(t²−3)²=(4−t²)(t²−1)²≥0. The maximum is 54+3·4=66, reached at (a,b,c)=(−1,−1,2). The tempting 78 comes from treating a², b², c² as if all three could equal 2; their square roots cannot then sum to zero.