If x,y,k are positive reals such that 3=k2(y2x2+x2y2)+k(yx+xy) find the maximum possible value of k.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
We have 3=k2(x2/y2+y2/x2)+k(x/y+y/x)≥2k2+2k, hence 7≥4k2+4k+1=(2k+1)2, hence k≤(7−1)/2. Obviously k can assume this value, if we let x=y=1.
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: Omni-MATH,
licensed Apache-2.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.