Olympiad Maths Prep

Track / Stage 6 / 222 of 400 #1222 of 2000

Problem 1222

National olympiad, first round
Algebra Difficulty 6.3 Prove it

9・202 Given non-negative real numbers x1,x2,⋯ ,xnx_{1}, x_{2}, \cdots, x_{n}. If x1+x2+⋯+xn=x_{1}+x_{2}+\cdots+x_{n}= nn, prove:
x11+x12+x21+x22+⋯+xn1+xn2⩽11+x1+11+x2+⋯+11+xn\begin{array}{c} \frac{x_{1}}{1+x_{1}^{2}}+\frac{x_{2}}{1+x_{2}^{2}}+\cdots+\frac{x_{n}}{1+x_{n}^{2}} \leqslant \frac{1}{1+x_{1}}+\frac{1}{1+x_{2}}+\cdots+ \\ \frac{1}{1+x_{n}} \end{array}

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

[Proof] For 1⩽k⩽n1 \leqslant k \leqslant n we have
xk1+xk2−11+xk=xk−1(1+xk2)(1+xk)\frac{x_{k}}{1+x_{k}^{2}}-\frac{1}{1+x_{k}}=\frac{x_{k}-1}{\left(1+x_{k}^{2}\right)\left(1+x_{k}\right)}

If xk⩾1x_{k} \geqslant 1, then (1+xk2)(1+xk)⩾4\left(1+x_{k}^{2}\right)\left(1+x_{k}\right) \geqslant 4. If xk<1x_{k}<1, then (1+xk2)(1+xk)<\left(1+x_{k}^{2}\right)\left(1+x_{k}\right)<
4. Thus, in any case we have
xk−1(1+xk2)(1+xk)⩽xk−14\frac{x_{k}-1}{\left(1+x_{k}^{2}\right)\left(1+x_{k}\right)} \leqslant \frac{x_{k}-1}{4}

Therefore, we get
∑k=1n(xk1+xk2−11+xk)⩽14∑k=1n(xk−1)=0\sum_{k=1}^{n}\left(\frac{x_{k}}{1+x_{k}^{2}}-\frac{1}{1+x_{k}}\right) \leqslant \frac{1}{4} \sum_{k=1}^{n}\left(x_{k}-1\right)=0

Thus, the original inequality holds.

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