Maths Olympiad Prep

Library / Taiwan IMO Selection Camp /260 of 359

Taiwan IMO Selection Camp

Algebra Difficulty 6.5 National Olympiad Prove it Taiwan

Let a,b,ca, b, c be non-negative real numbers such that (a+b)(b+c)(c+a)≠0(a+b)(b+c)(c+a) \neq 0. Find the minimum of
(a+b+c)2016(1a2016+b2016+1b2016+c2016+1c2016+a2016). (a + b + c)^{2016} \left( \frac{1}{a^{2016} + b^{2016}} + \frac{1}{b^{2016} + c^{2016}} + \frac{1}{c^{2016} + a^{2016}} \right).

Solution

When xy+yz+zx=1xy + yz + zx = 1 and (x+y)(y+z)(z+x)≠0(x + y)(y + z)(z + x) \neq 0, we have
1x+y+1y+z+1z+x≥52.(1) \frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} \ge \frac{5}{2}. \qquad (1)
Now, WLOG we can assume a2016b2016+b2016c2016+c2016a2016=1a^{2016}b^{2016}+b^{2016}c^{2016}+c^{2016}a^{2016} = 1. Then, we have
1a2016+b2016+1b2016+c2016+1c2016+a2016≥52.(2) \frac{1}{a^{2016} + b^{2016}} + \frac{1}{b^{2016} + c^{2016}} + \frac{1}{c^{2016} + a^{2016}} \ge \frac{5}{2}. \qquad (2)
Furthermore, by Muirhead and Schur's inequality of degree 3, we have
(a+b+c)4032≥(a4032+b4032+c4032+3a1344+b1344+c1344)+∑i=12015Ci2016(aib2016−i+bic2016−i+cia2016−i)≥24032(a2016b2016+b2016c2016+c2016a2016)=24032. (a+b+c)^{4032} \\ \ge (a^{4032} + b^{4032} + c^{4032} + 3a^{1344} + b^{1344} + c^{1344}) + \sum_{i=1}^{2015} C_i^{2016} (a^i b^{2016-i} + b^i c^{2016-i} + c^i a^{2016-i}) \\ \ge 2^{4032} (a^{2016}b^{2016} + b^{2016}c^{2016} + c^{2016}a^{2016}) = 2^{4032}.
Combining these, we get the minimum value is 5×220155 \times 2^{2015}.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.