JOHN
Lv 7
JOHN asked in Science & MathematicsMathematics · 3 years ago

If a > 0, b > 0, c > 0, prove that (a² + 2)(b² + 2)(c² + 2) ≥ 9(ab + bc + ca).?

3 Answers

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  • Indica
    Lv 7
    3 years ago
    Favourite 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

    https://mks.mff.cuni.cz/kalva/apmo/asoln/asol045.h...

  • 3 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....
  • JOHN
    Lv 7
    3 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.

    Attachment image
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