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Algebra Difficulty 5.7 AIME, harder Prove it

26. Proof:
x1x2⋯xn=1,xi⩾0x_{1} x_{2} \cdots x_{n}=1, x_{i} \geqslant 0

implies x1+x2+⋯+xn⩾nx_{1}+x_{2}+\cdots+x_{n} \geqslant n.

Solution

26. Assuming the conclusion holds for nn, and
x1x2⋯xnxn+1=1x_{1} x_{2} \cdots x_{n} x_{n+1}=1

Without loss of generality, assume x1⩾1,x2⩽1x_{1} \geqslant 1, x_{2} \leqslant 1, then we have (x1−1)(x2−1)⩽0\left(x_{1}-1\right)\left(x_{2}-1\right) \leqslant 0, or
x1x2+1⩽x1+x2x_{1} x_{2}+1 \leqslant x_{1}+x_{2}

Therefore, applying the assumption to nn quantities x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n}, we get
x1+x2+x3+⋯+xn+1⩾1+x1x2+x3+⋯+xn+1⩾1+n.x_{1}+x_{2}+x_{3}+\cdots+x_{n+1} \geqslant 1+x_{1} x_{2}+x_{3}+\cdots+x_{n+1} \geqslant 1+n .

Since the conclusion is trivial for n=1n=1, the proposition 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.