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experimentX Group Title

If \( abc = 1 \) where \( a, b, c > 0 \) show that \[ a^2 + b^2+c^2 \ge a+b+c \] No Lagrange multipliers allowed.

  • one year ago
  • one year ago

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  1. wio Group Title
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    I think it involve multiplying a side by \(abc\)

    • one year ago
  2. experimentX Group Title
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    i just solved it after posting .. it was clever!! I'll keep the question open for a while.

    • one year ago
  3. experimentX Group Title
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    @wio would you like a hint .. since i had it.

    • one year ago
  4. wio Group Title
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    What's the hint?

    • one year ago
  5. experimentX Group Title
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    it's AM-GM inequality. Perhaps you are familiar. \[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \ge 3(ab+bc+ca) \] This is half of it, the other one, i presume you know it.

    • one year ago
  6. experimentX Group Title
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    sorry, also i forgot to specify that \( a, b, c > \) at the beginning.

    • one year ago
  7. wio Group Title
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    They're greater than 0? cause that changes a lot

    • one year ago
  8. experimentX Group Title
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    Well, the problem is stated as "Let a; b; c be positive real numbers such that abc = 1. Prove that ... "

    • one year ago
  9. wio Group Title
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    Ah ok.

    • one year ago
  10. wio Group Title
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    I was messing around with algebra getting dead ends like \[ \begin{array}{rcl} a^2 + b^2+c^2 &\ge& a+b+c \\ a^2 + b^2+c^2 &\ge& a^2bc+b^2ac+c^2ab \\ 0 &\ge& (bc-1)+(ac-1)+(ab-1) \\ 0 &\ge& bc+ac+ab-3 \\ 3 &\ge& bc+ac+ab \\ \end{array} \]

    • one year ago
  11. experimentX Group Title
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    Here's a short proof hint ... i had written earlier. \[ \begin{align*} 3(a^2+b^2+c^2) & = a^2 + b^2 + c^2 + (a^2+b^2) + (b^2 + c^2) + (c^2+a^2) \\ & \ge a^2 + b^2 +c^2 + 2 a b + 2 b c + 2 c a \\ & = (a+b+c)^2 \\ & = \frac{2(a^2 + b^2 +c^2)}{2} + 2 a b + 2 b c + 2 ca \\ & \ge \frac{ 2 a b + 2 b c + 2 ca }{2} + 2 a b + 2 b c + 2 ca \\ & = 3(a+b+c) \end{align*} \]

    • one year ago
  12. experimentX Group Title
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    the other hint would to be make use of AM-GM inequality for 3 variables.

    • one year ago
  13. experimentX Group Title
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    \[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \\ 3 = 3 \sqrt[3]{abc} \le (a+b+c) \] Dividing will do the job!!

    • one year ago
  14. sauravshakya Group Title
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    (a+b+c)/3 >= (abc)^1/3 a+b+c>=3 Now, a^2+b^2+c^2=(a+b+c)^2 -2ab -2bc-ca SO, a^2+b^2+c^2>=(a+b+c)^2 .....i Since, a+b+c>=3 (a+b+c)^2 >=a+b+c ....ii Thus, from i and ii, a^2+b^2+c^2 >=a+b+c

    • one year ago
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