Maths Olympiad Prep

Library / /509 of 520

Algebra Difficulty 6.0 National olympiad Prove it

Let a,b,ca, b, c be positive numbers, and ∑\sum denote the cyclic sum over the relevant quantities. Then for n⩾1n \geqslant 1 we have
∑anbn+cn⩾∑an−1bn−1+cn−1,∑an+1bn+cn⩾∑anbn−1+cn−1. \begin{array}{l} \sum \frac{a^{n}}{b^{n}+c^{n}} \geqslant \sum \frac{a^{n-1}}{b^{n-1}+c^{n-1}}, \\ \sum \frac{a^{n+1}}{b^{n}+c^{n}} \geqslant \sum \frac{a^{n}}{b^{n-1}+c^{n-1}} . \end{array}

Solution

Proof: (1) Without loss of generality, let a⩾b⩾ca \geqslant b \geqslant c, then it is easy to prove
anbn−1+ancn−1−an−1bn−an−1cn⩾0,ancn−1−an−1cn+bncn−1−bn−1cn⩾0, \begin{array}{l} a^{n} b^{n-1}+a^{n} c^{n-1}-a^{n-1} b^{n}-a^{n-1} c^{n} \geqslant 0, \\ a^{n} c^{n-1}-a^{n-1} c^{n}+b^{n} c^{n-1}-b^{n-1} c^{n} \geqslant 0, \end{array}

and
+bnan−1+bncn−1−bn−1an−bn−1cn(an+cn)(an−1+cn−1)⩾anbn−1+ancn−1−an−1bn−an−1cn(an+cn)(an−1+cn−1)+bnan−1+bncn−1−bn−1an−bn−1cn(an+cn)(an−1+cn−1)=ancn−1−an−1cn+bncn−1−bn−1cn(an+cn)(an−1+cn−1)⩾ancn−1−an−1cn+bncn−1−bn−1cn(an+bn)(an−1+bn−1)=cn−1an−1+bn−1−cnan+bn.  \begin{array}{l} +\frac{b^{n} a^{n-1}+b^{n} c^{n-1}-b^{n-1} a^{n}-b^{n-1} c^{n}}{\left(a^{n}+c^{n}\right)\left(a^{n-1}+c^{n-1}\right)} \\ \geqslant \frac{a^{n} b^{n-1}+a^{n} c^{n-1}-a^{n-1} b^{n}-a^{n-1} c^{n}}{\left(a^{n}+c^{n}\right)\left(a^{n-1}+c^{n-1}\right)} \\ +\frac{b^{n} a^{n-1}+b^{n} c^{n-1}-b^{n-1} a^{n}-b^{n-1} c^{n}}{\left(a^{n}+c^{n}\right)\left(a^{n-1}+c^{n-1}\right)} \\ =\frac{a^{n} c^{n-1}-a^{n-1} c^{n}+b^{n} c^{n-1}-b^{n-1} c^{n}}{\left(a^{n}+c^{n}\right)\left(a^{n-1}+c^{n-1}\right)} \\ \geqslant \frac{a^{n} c^{n-1}-a^{n-1} c^{n}+b^{n} c^{n-1}-b^{n-1} c^{n}}{\left(a^{n}+b^{n}\right)\left(a^{n-1}+b^{n-1}\right)} \\ =\frac{c^{n-1}}{a^{n-1}+b^{n-1}}-\frac{c^{n}}{a^{n}+b^{n}} \text {. } \\ \end{array}

At this point, inequality (1) holds.
(2) The proof of inequality (2) is extremely similar to that of inequality (1), and is omitted.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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