no..
simple bracket
Find largest possible integer n such that [\frac{na^2b^2c^2}{a^2bc+ab^2c+abc^2}]\leq (a+b)^2+(a+b+4c)^2
for all real positive a,b,c
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4 Answers
Mani Pal Singh
·2009-08-18 11:11:28
nabca+b+c ≤ 2a2+2b2+16c2+4ab+8bc+8ac
nabc≤2a3+2b3+16c3+4a2b+20abc+8a2c+4ab2+8b2c+8bc2+8ac2
≤-nabc+2(a+b+2c)3-............................... (1)
RHS ≥ 0
Solve (1)