A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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.
 2 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.

This Question is Closed

wio
 2 years ago
Best ResponseYou've already chosen the best response.1I think it involve multiplying a side by \(abc\)

experimentX
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1@wio would you like a hint .. since i had it.

experimentX
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1sorry, also i forgot to specify that \( a, b, c > \) at the beginning.

wio
 2 years ago
Best ResponseYou've already chosen the best response.1They're greater than 0? cause that changes a lot

experimentX
 2 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 ... "

wio
 2 years ago
Best ResponseYou've already chosen the best response.1I 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
 2 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
 2 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
 2 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!!

sauravshakya
 2 years ago
Best 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
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.