Math Olympiad Question | Inequality | Cauchy-Schwarz Inequality | Algebra

How to solve this Mathematical Olympiad problem? How to prove this inequality? This algebra video will show you the techniques and tricks like Cauchy-Schwarz inequality to solve this challenging Math Olympiad question. For another Math Olympiad video: • Math Olympiad Question...
For more questions, tips and tricks like this, please visit Dr Wang's Channel: / @drwangusa
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[@nicolascamargo8339]

Wow excelente aplicación de ese teorema

[@ranshen1486]

I'm curious in a tight bound. Introducing free coefs {a,b,c} to the LHS, I can rewrite it as LHS = [1/√a] * √(4ax+4a) + [1/√b] * √(2bx-3b) + [1/√c] * √(15c-3cx). Cauchy-Schwarz inequality will be applied with parts in [ ] as one vector and the √ part outside [ ] as the other. So LHS² ≤ (1/a + 1/b + 1/c) { (4a + 2b - 3c)x + 4a - 3b + 15c }. In order to cancel all "x" term, we need to set its coef to 0, (4a + 2b - 3c) = 0, i.e., c = (4a + 2b)/3. After removing c, I got {1/a + 1/b + 3/(4a+2b) } * { 24a + 7b}. Since any common factor of both a and b will cancel out in the above expression, we can WLOG let R = b/a and a=1. After some trivial algebra, I got (1 + 1/R + 3/(4+2R)) (24+7R) = (14R³ + 111R² + 244R +96) / (2R² + 4R) as the bound, which I need to minimize over R > 0. Setting its derivative to 0, I got 7R⁴ + 28R³ - 11R² - 96R - 96 = 0. Unfortunately, I couldn't find any simple solution for this quartic (although I know it is there, it's just too messy to me). However, the only positive real root is approximately R ≈ 1.99. To make it nicer to work with, I will take R=2 in order to find a tigher upper bound. With a=1, b=R=2, and c=8/3, the LHS of the inequality is rewritten as √(4x+4) + [1/√2] * √(4x - 6) + [√(3/8)] * √(40 - 8x), and Caucy-Schwartz inequality gives LHS² < { 1 + 1/2 + 3/8 } * { 4 - 6 + 40 } = 71¼ which is tighter than 76=2²×19 in the question, as expected. If I use the numerical solution R = 1.9904, the maximum is √71.2497, only slightly below √71¼ given above.