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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it involve multiplying a side by \(abc\)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1i just solved it after posting .. it was clever!! I'll keep the question open for a while.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1@wio would you like a hint .. since i had it.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1it's AMGM 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.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1sorry, also i forgot to specify that \( a, b, c > \) at the beginning.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0They're greater than 0? cause that changes a lot

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1Well, the problem is stated as "Let a; b; c be positive real numbers such that abc = 1. Prove that ... "

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was messing around with algebra getting dead ends like \[ \begin{array}{rcl} a^2 + b^2+c^2 &\ge& a+b+c \\ a^2 + b^2+c^2 &\ge& a^2bc+b^2ac+c^2ab \\ 0 &\ge& (bc1)+(ac1)+(ab1) \\ 0 &\ge& bc+ac+ab3 \\ 3 &\ge& bc+ac+ab \\ \end{array} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1Here's a short proof hint ... i had written earlier. \[ \begin{align*} 3(a^2+b^2+c^2) & = a^2 + b^2 + c^2 + (a^2+b^2) + (b^2 + c^2) + (c^2+a^2) \\ & \ge a^2 + b^2 +c^2 + 2 a b + 2 b c + 2 c a \\ & = (a+b+c)^2 \\ & = \frac{2(a^2 + b^2 +c^2)}{2} + 2 a b + 2 b c + 2 ca \\ & \ge \frac{ 2 a b + 2 b c + 2 ca }{2} + 2 a b + 2 b c + 2 ca \\ & = 3(a+b+c) \end{align*} \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1the other hint would to be make use of AMGM inequality for 3 variables.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ 3(a^2+b^2+c^2) \ge (a+b+c)^2 \\ 3 = 3 \sqrt[3]{abc} \le (a+b+c) \] Dividing will do the job!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(a+b+c)/3 >= (abc)^1/3 a+b+c>=3 Now, a^2+b^2+c^2=(a+b+c)^2 2ab 2bcca 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
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