Olympiad Maths Prep

Track / Stage 7 / 140 of 300 #1540 of 2000

Problem 1540

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.3 Prove it

4. Prove that for any α⩽1\alpha \leqslant 1 and any real numbers x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n} satisfying 1⩾x1⩾x2⩾⋯⩾xn>01 \geqslant x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}>0, we have (1+x1+x2+⋯+xn)α⩽1+1α−1x1α+2α−1x2α+⋯+nα−1xnα\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha} \leqslant 1+1^{\alpha-1} x_{1}^{\alpha}+2^{\alpha-1} x_{2}^{\alpha}+\cdots+n^{\alpha-1} x_{n}^{\alpha}.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

4. When n=Fn=\mathrm{F}, (1+x1)α⩽1+x1α\left(1+x_{1}\right)^{\alpha} \leqslant 1+x_{1}^{\alpha} (this can be obtained by using the fact that f(x)=(1+x)α−(1+xα)f(x)=(1+x)^{\alpha}-\left(1+x^{\alpha}\right) is a decreasing function on (0,1)(0,1)), the inequality holds. Assume the inequality holds for nn, we will prove it also holds for n+1n+1. We have
(1+xF+x2+⋯+xn+xn+1)α−(F+x1+x2+⋯+xn)α=(1+x1+x2+⋯+xn)α[(F+1+x1+x2+⋯+xn)α−F]⩽(1+x1+x2+⋯+xn)αxn+1⩽((n+1)xn+1)α−1xn+1=(n+1)α−1xn+1α\begin{array}{l} \left(1+x_{\mathrm{F}}+x_{2}+\cdots+x_{n}+x_{n+1}\right)^{\alpha}-\left(\mathrm{F}+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha}= \\ \left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha}\left[\left(\mathrm{F}+1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha}-\mathrm{F}\right] \leqslant \\ \left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha} x_{n+1} \leqslant \\ \left((n+1) x_{n+1}\right)^{\alpha-1} x_{n+1}=(n+1)^{\alpha-1} x_{n+1}^{\alpha} \end{array}

The inequality (1+x1+x2+⋯+xn)α−1xn+1⩽((n+1)n+1)α−1xn+1\left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha-1} x_{n+1} \leqslant\left((n+1)_{n+1}\right)^{\alpha-1} x_{n+1}
is valid by the condition 1+x1+x2+⋯+xn⩾(n+1)xn+11+x_{1}+x_{2}+\cdots+x_{n} \geqslant(n+1) x_{n+1}
and α−1⩽0\alpha-1 \leqslant 0. By the proven inequality and the induction hypothesis, we get
(1+x1+x2+⋯+xn+xn+1)α⩽(1+x1+x2+⋯+xn)α+(−n+1)α−1xn+1α⩽1+1α−1x1α+2α−1x2α+⋯+nα−1xnα+(n+1)α−1xn+1α\begin{array}{l} \left(1+x_{1}+x_{2}+\cdots+x_{n}+x_{n+1}\right)^{\alpha} \leqslant \\ \left(1+x_{1}+x_{2}+\cdots+x_{n}\right)^{\alpha}+(-n+1)^{\alpha-1} x_{n+1}^{\alpha} \leqslant \\ 1+1^{\alpha-1} x_{1}^{\alpha}+2^{\alpha-1} x_{2}^{\alpha}+\cdots+n^{\alpha-1} x_{n}^{\alpha}+(n+1)^{\alpha-1} x_{n+1}^{\alpha} \end{array}

This is what we need to prove.

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