Let a, b, c be positive real numbers such that a 1 + a + b 1 + b + c 1 + c = 1. prove that abc ≤ 1/8.
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This is how I tried doing it:
Let p=a+b+c,p=a+b+c, and q=1a+1b+1cq=1a+1b+1c.
Using AM>GM for p,qp,q, I get:
p+q2>(pq)1/2p+q2>(pq)1/2
(a+b+c)(1a+1b+1c)−−−−−−−−−−−−−−−−−−−−√<(a+1a+b+1b+c+1c)2(a+b+c)(1a+1b+1c)<(a+1a+b+1b+c+1c)2
And for any x∈R, x+1x≥2.x∈R, x+1x≥2.
Thus,
(a+b+c)(1a+1b+1c)<9,(a+b+c)(1a+1b+1c)<9,
Let p=a+b+c,p=a+b+c, and q=1a+1b+1cq=1a+1b+1c.
Using AM>GM for p,qp,q, I get:
p+q2>(pq)1/2p+q2>(pq)1/2
(a+b+c)(1a+1b+1c)−−−−−−−−−−−−−−−−−−−−√<(a+1a+b+1b+c+1c)2(a+b+c)(1a+1b+1c)<(a+1a+b+1b+c+1c)2
And for any x∈R, x+1x≥2.x∈R, x+1x≥2.
Thus,
(a+b+c)(1a+1b+1c)<9,(a+b+c)(1a+1b+1c)<9,
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