## experimentX 3 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.

1. anonymous

I think it involve multiplying a side by $$abc$$

2. experimentX

i just solved it after posting .. it was clever!! I'll keep the question open for a while.

3. experimentX

@wio would you like a hint .. since i had it.

4. anonymous

What's the hint?

5. experimentX

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.

6. experimentX

sorry, also i forgot to specify that $$a, b, c >$$ at the beginning.

7. anonymous

They're greater than 0? cause that changes a lot

8. experimentX

Well, the problem is stated as "Let a; b; c be positive real numbers such that abc = 1. Prove that ... "

9. anonymous

Ah ok.

10. anonymous

I was messing around with algebra getting dead ends like $\begin{array}{rcl} a^2 + b^2+c^2 &\ge& a+b+c \\ a^2 + b^2+c^2 &\ge& a^2bc+b^2ac+c^2ab \\ 0 &\ge& (bc-1)+(ac-1)+(ab-1) \\ 0 &\ge& bc+ac+ab-3 \\ 3 &\ge& bc+ac+ab \\ \end{array}$

11. experimentX

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*}

12. experimentX

the other hint would to be make use of AM-GM inequality for 3 variables.

13. experimentX

$3(a^2+b^2+c^2) \ge (a+b+c)^2 \\ 3 = 3 \sqrt[3]{abc} \le (a+b+c)$ Dividing will do the job!!

14. anonymous

(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