Let a1,a2,…,an be positive real numbers such that a1+a2+⋯+an=a12+a22+⋯+an2. Prove that 1≤i<j≤n∑aiaj(1−aiaj)≥0.
Solution
The original inequality is equivalent to 21≤i<j≤n∑aiaj≥21≤i<j≤n∑ai2aj2, or (i=1∑nai)2−i=1∑nai2≥(i=1∑nai2)2−i=1∑nai4. Using the given hypothesis, it can be rewritten as a14+a24+⋯+an4≥a12+a22+⋯+an2. To prove this last inequality, we have from Holder's inequality (a14+a24+⋯+an4)(a1+a2+⋯+an)2≥(a12+a22+⋯+an2)3. Using the given hypothesis again, we get the result.
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 and solution reproduced as published; topic and difficulty added by this site.