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Home » a + b + c \geq 3\sqrt[3]abc = 3. - go-checkin.com

a + b + c \geq 3\sqrt[3]abc = 3. - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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= (A + B)k + A
For the equation to hold for all \(k\), we set up the system:
A + B = 0 \quad \text{and} \quad A = 1
Solving this, we find \(A = 1\) and \(B = -1\). Thus:
Now, substitute this into the sum:
This is a telescoping series, where most terms cancel out:
After cancellation, we are left with:
Thus, the sum is:
- \frac{1}{51} = \frac{51}{51} - \frac{1}{51} = \frac{50}{51}
The final answer is:

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