Maths Olympiad Prep

Library / /20 of 28

Algebra Difficulty 4.9 AIME Prove it Philippines

Problem:
Find the smallest number kk such that for all real numbers xx, yy and zz
(x2+y2+z2)2k(x4+y4+z4) \left(x^{2}+y^{2}+z^{2}\right)^{2} \leq k\left(x^{4}+y^{4}+z^{4}\right)

Solution

Solution:
Note that
(x2+y2+z2)2=x4+y4+z4+2x2y2+2x2z2+2y2z2 \left(x^{2}+y^{2}+z^{2}\right)^{2}=x^{4}+y^{4}+z^{4}+2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2}
Using the AM-GM Inequality, we find that
2x2y2+2x2z2+2y2z22[(x4+y4)/2]+2[(x4+z4)/2]+2[(y4+z4)/2] 2 x^{2} y^{2}+2 x^{2} z^{2}+2 y^{2} z^{2} \leq 2\left[\left(x^{4}+y^{4}\right) / 2\right]+2\left[\left(x^{4}+z^{4}\right) / 2\right]+2\left[\left(y^{4}+z^{4}\right) / 2\right]
which when combined with the previous equation results to
(x2+y2+z2)23(x4+y4+z4) \left(x^{2}+y^{2}+z^{2}\right)^{2} \leq 3\left(x^{4}+y^{4}+z^{4}\right)
Therefore k=3k=3.

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: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.