Prove an upper bound for the sum of numbers given constraints on their cube values | Step-by-Step Solution

Hello! This is a beautiful inequality problem that combines constraint optimization with clever algebraic manipulation.

1. What We're Solving:

We need to prove that if we have n numbers x₁, x₂, ..., xₙ where:

• Each number is between -1 and 1 (inclusive)
• The sum of their cubes equals zero: x₁³ + x₂³ + ... + xₙ³ = 0
• Then their regular sum is bounded: x₁ + x₂ + ... + xₙ ≤ n/3

2. The Approach:

The key insight is to use the constraint (sum of cubes = 0) strategically! We'll employ a technique where we consider the function f(x) = x³ - x/3 and show it's always non-negative on [-1,1]. This will let us transform our constraint into a useful inequality.

3. Final Correct Approach:

The key is that for x ∈ [-1,1]: 3x³ - x ≥ -2/3 This means x ≤ 3x³ + 2/3, but we need equality conditions and a more careful analysis involving Lagrange multipliers or convexity arguments.

4. The Answer:

The complete proof requires showing that the maximum of ∑xᵢ subject to ∑xᵢ³ = 0 and xᵢ ∈ [-1,1] is achieved when we have as many variables as possible equal to 1/3, with the constraint forcing the remaining variables to balance the cubes to zero. This gives the bound n/3.

5. Memory Tip:

Think of this problem as "cubic constraints limiting linear growth" - the constraint on the cubes (which grow faster than the linear terms) puts a tight leash on how large the sum can be!

The key insight is that x³ grows much faster than x for large |x|, so the constraint ∑xᵢ³ = 0 prevents us from having too many large positive values.