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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
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

This Question is Closed

wio Group TitleBest ResponseYou've already chosen the best response.1
I think it involve multiplying a side by \(abc\)
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
@wio would you like a hint .. since i had it.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
What's the hint?
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
it'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.
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.1
They're greater than 0? cause that changes a lot
 one year ago

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

wio Group TitleBest ResponseYou've already chosen the best response.1
I was messing around with algebra getting dead ends like \[ \begin{array}{rcl} a^2 + b^2+c^2 &\ge& a+b+c \\ a^2 + b^2+c^2 &\ge& a^2bc+b^2ac+c^2ab \\ 0 &\ge& (bc1)+(ac1)+(ab1) \\ 0 &\ge& bc+ac+ab3 \\ 3 &\ge& bc+ac+ab \\ \end{array} \]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
the other hint would to be make use of AMGM inequality for 3 variables.
 one year ago

experimentX Group TitleBest 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!!
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.1
(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
 one year ago
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