### 3 Answers

- IndicaLv 73 years agoFavourite answer
Expand to get 8 + 4∑a² + 2∑a²b² + a²b²c² − 9∑ab ≥ 0 … (i)

Then use uvw method subs 3u=a+b+c, 3v=ab+bc+ca, w=abc (so there is equality at u=v=w=1)

∑a² = (∑a)²−2∑ab = 9u²−6v

∑a²b² = (∑ab)²−2∑abbc = (∑ab)²−abc∑a = 9v²−6uw

(i) is now 8 + 4(9u²−6v) + 2(9v²−6uw) + w² − 27v ≥ 0 or 8+36u²+18v²−12uw+w²−51v ≥ 0

LHS can be expressed as a sum of positive terms where each component is zero at u=v=w=1 !

(w−u)² + 8(v−1)² + 35(u²−v) + 10(v²−uw)

First two are obviously positive and the last two follow from (x+y+z)² ≥ 3(xy+yz+zx)

Set x=a, y=b, z=c for (∑a)² ≥ 3∑ab ⟹ 9u²−9v ≥ 0

Set x=ab, y=bc, z=ac for (∑ab)² ≥ 3∑abbc or (∑ab)² ≥ 3abc∑a ⟹ 9v²−9uw ≥ 0

∴ (w−u)² + 8(v−1)² + 35(u²−v) + 10(v²−uw) ≥ 0

Also see

- Saurabh DubeyLv 43 years ago
• (ab-1)² + (bc-1)² +(ca-1)² ≥0

(a²b² + b²c² + c²a²) + 3 ≥ 2(ab + bc + ca )

2(a²b² + b²c² + c²a²) + 6 ≥ 4(ab + bc + ca )....(1)

• (a - b)² ≥ 0

a² + b² ≥ 2ab......(2)

similarly,

b² + c² ≥ 2bc.......(3)

c² + a² ≥ 2ca........(4)

Adding (2), (3) and (4)

2( a² + b² + c²) ≥ 2(ab + bc + ca )

3 (a² + b² + c²) ≥ 3(ab + bc + ca )........(5)

••••(a² + b² + c² ) + (abc)² + 2 ≥ 2(ab + bc + ca)

..........(6)

Adding equation (1) (5) and (6)

2(a²b² + b²c² + c²a²) + 4(a² + b² + c² ) + (abc)²

+ 8 ≥ 9 ( ab + bc + ca )

i.e

(a² + 2) (b² + 2) (c² + 2) ≥ 9(ab + bc + ca )

Source(s): •For equation (6) https://in.answers.yahoo.com/question/index?qid=20... •You have already proved equation (6). •I had posted that question. •I was stucked in that , I will try to find other method to prove equation (6) without using any specific theorem ... later.... - JOHNLv 73 years ago
Solution

EDIT: Think I've found a non-calculus solution. Too tired now (1:20am UK time) to check it - will do so tomorrow. Chances are I've made a mistake somewhere. But I'll check it Wednesday.