Maths Olympiad Prep

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Algebra Difficulty 5.7 AIME, harder Prove it India

Problem:

Let n≥3n \geq 3 be an integer and let 1<a1≤a2≤a3≤⋯≤an1 < a_{1} \leq a_{2} \leq a_{3} \leq \cdots \leq a_{n} be nn real numbers such that a1+a2+a3+⋯+an=2na_{1} + a_{2} + a_{3} + \cdots + a_{n} = 2n. Prove that
a1a2⋯an−1+a1a2⋯an−2+⋯+a1a2+a1+2≤a1a2⋯an a_{1} a_{2} \cdots a_{n-1} + a_{1} a_{2} \cdots a_{n-2} + \cdots + a_{1} a_{2} + a_{1} + 2 \leq a_{1} a_{2} \cdots a_{n}

Solution

Solution:

We use Chebyshev's inequality. Observe
n(a1a2⋯an−1+a1a2⋯an−2+⋯+a1+1)=(a1a2⋯an−1+a1a2⋯an−2+⋯+a1+1)((an−1)+(an−1−1)+⋯+(a1−1))≤n(a1a2⋯an−1(an−1)+⋯+a1(a2−1)+1(a1−1))≤n(a1a2⋯an−1) \begin{aligned} & n\left(a_{1} a_{2} \cdots a_{n-1} + a_{1} a_{2} \cdots a_{n-2} + \cdots + a_{1} + 1\right) \\ & \quad = \left(a_{1} a_{2} \cdots a_{n-1} + a_{1} a_{2} \cdots a_{n-2} + \cdots + a_{1} + 1\right)\left(\left(a_{n} - 1\right) + \left(a_{n-1} - 1\right) + \cdots + \left(a_{1} - 1\right)\right) \\ & \quad \leq n\left(a_{1} a_{2} \cdots a_{n-1}\left(a_{n} - 1\right) + \cdots + a_{1}\left(a_{2} - 1\right) + 1\left(a_{1} - 1\right)\right) \\ & \quad \leq n\left(a_{1} a_{2} \cdots a_{n} - 1\right) \end{aligned}
It follows that
a1a2⋯an−1+a1a2⋯an−2+⋯+a1+1≤a1a2⋯an−1 a_{1} a_{2} \cdots a_{n-1} + a_{1} a_{2} \cdots a_{n-2} + \cdots + a_{1} + 1 \leq a_{1} a_{2} \cdots a_{n} - 1
This gives the required inequality.

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