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Prove an inequality abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)
- Thread starter naly
- Start date
It would be the same situation as multiplying each side by 3:
3abc(a+b+c)2 ≤ 3(a3+b3+c3)(ab+bc+ca)
You stated that you know that 3abc ≤ (a3+b3+c3).
Then, what about showing \(\displaystyle \ (a + b + c)^2 \le 3(ab + bc + ca) \ \)?
thank you, but I just proved an opposite answer.
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
for ab+bc+ca ≤ a2+b2+c2
I got 3ab+3bc+3ca ≤ a2+b2+c2+2ab+2bc+2ca
so 3(ab+bc+ca) ≤ (a+b+c)2
is there something wrong?
D
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You state:thank you, but I just proved an opposite answer.
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
for ab+bc+ca ≤ a2+b2+c2
I got 3ab+3bc+3ca ≤ a2+b2+c2+2ab+2bc+2ca
so 3(ab+bc+ca) ≤ (a+b+c)2
is there something wrong?
for ab+bc+ca ≤ a2+b2+c2
Is that a relationship given to you or you had derived it.I derived it from a2+b2+c2-(ab+bc+ca) = 1/2{(a-b)2+(b-c)2+(c-a)2}You state:
for ab+bc+ca ≤ a2+b2+c2Is that a relationship given to you or you had derived it.
Hey, I don't know if you already managed to solve it or if you'd still like to see the solution, but I casually found this post and I thought I might as well respond it.
Your approach is intuitive and that was also the first thing I thought; however, there is a problem when trying to prove (a+b+c)^2 ≤ 3(ab+bc+ca), because, in fact, the opposite is true: (a+b+c)^2 ≥ 3(ab+bc+ca). You can see that if you expand (a+b+c)^2, simplify, multiply by 2, and use the trivial inequality.
Instead of doing AM-GM, I managed to solve it using Cauchy-Schwartz Inequality. There is likely a solution with AM-GM, but I don't see it. So here is mine:
(a3+b3+c3)(ab+bc+ca) ≥abc(a+b+c)2
Divide all by abc
(a3+b3+c3)(1/a+1/b+1/c) ≥(a+b+c)2
Now, the LHS = (a3+b3+c3)(1/a+1/b+1/c) = (sqrt(a3)2+sqrt(b3)2+sqrt(c3)2)(sqrt(1/a)2+sqrt(1/b)2+sqrt(1/c)2)
Then, by Cauchy-Shwartz, the LHS ≥ (sqrt(a3/a) + sqrt(b3/b) + sqrt(c3/c))2 = (a+b+c)2
Which proves the original inequality.
Your approach is intuitive and that was also the first thing I thought; however, there is a problem when trying to prove (a+b+c)^2 ≤ 3(ab+bc+ca), because, in fact, the opposite is true: (a+b+c)^2 ≥ 3(ab+bc+ca). You can see that if you expand (a+b+c)^2, simplify, multiply by 2, and use the trivial inequality.
Instead of doing AM-GM, I managed to solve it using Cauchy-Schwartz Inequality. There is likely a solution with AM-GM, but I don't see it. So here is mine:
(a3+b3+c3)(ab+bc+ca) ≥abc(a+b+c)2
Divide all by abc
(a3+b3+c3)(1/a+1/b+1/c) ≥(a+b+c)2
Now, the LHS = (a3+b3+c3)(1/a+1/b+1/c) = (sqrt(a3)2+sqrt(b3)2+sqrt(c3)2)(sqrt(1/a)2+sqrt(1/b)2+sqrt(1/c)2)
Then, by Cauchy-Shwartz, the LHS ≥ (sqrt(a3/a) + sqrt(b3/b) + sqrt(c3/c))2 = (a+b+c)2
Which proves the original inequality.