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Algebra Difficulty 5.8 AIME, harder Prove it

1. Let nn be a positive integer, and a1,a2,⋯ ,ana_{1}, a_{2}, \cdots, a_{n} be non-negative real numbers. Prove:
11+a1+a1(1+a1)(1+a2)+⋯+a1a2⋯an−1(1+a1)(1+a2)⋯(1+an)⩽1. \begin{array}{l} \frac{1}{1+a_{1}}+\frac{a_{1}}{\left(1+a_{1}\right)\left(1+a_{2}\right)}+\cdots+ \\ \frac{a_{1} a_{2} \cdots a_{n-1}}{\left(1+a_{1}\right)\left(1+a_{2}\right) \cdots\left(1+a_{n}\right)} \leqslant 1 . \end{array}

Solution

1. Notice that, 11+a=1−a1+a\frac{1}{1+a}=1-\frac{a}{1+a}.

Therefore, for any i(i=1,2,⋯ ,n)i(i=1,2, \cdots, n) we have
a11+a1⋯⋅ai−11+ai−1⋅11+ai=a11+a1⋯⋅ai−11+ai−1(1−ai1+ai)=a11+a1⋯⋯ai−11+ai−1−a11+a1⋯⋯ai1+ai. \begin{array}{l} \frac{a_{1}}{1+a_{1}} \cdots \cdot \frac{a_{i-1}}{1+a_{i-1}} \cdot \frac{1}{1+a_{i}} \\ =\frac{a_{1}}{1+a_{1}} \cdots \cdot \frac{a_{i-1}}{1+a_{i-1}}\left(1-\frac{a_{i}}{1+a_{i}}\right) \\ =\frac{a_{1}}{1+a_{1}} \cdots \cdots \frac{a_{i-1}}{1+a_{i-1}}-\frac{a_{1}}{1+a_{1}} \cdots \cdots \frac{a_{i}}{1+a_{i}} . \end{array}

Summing all these equations, we get
∑i=1na11+a1⋯⋯ai−11+ai−1⋅11+ai=∑i=1n(a11+a1⋯⋅ai−11+ai−1−a11+a1⋯⋯ai1+ai)=1−a11+a1⋯⋯an1+an. \begin{array}{l} \sum_{i=1}^{n} \frac{a_{1}}{1+a_{1}} \cdots \cdots \frac{a_{i-1}}{1+a_{i-1}} \cdot \frac{1}{1+a_{i}} \\ =\sum_{i=1}^{n}\left(\frac{a_{1}}{1+a_{1}} \cdots \cdot \frac{a_{i-1}}{1+a_{i-1}}-\frac{a_{1}}{1+a_{1}} \cdots \cdots \frac{a_{i}}{1+a_{i}}\right) \\ =1-\frac{a_{1}}{1+a_{1}} \cdots \cdots \frac{a_{n}}{1+a_{n}} . \end{array}

Thus, the inequality holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.