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Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.

My (strange) proof:

$$ \begin{align*} a^3+b^3+c^3 &\geq a^2b+b^2c+c^2a\\ \sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a^2 &\geq \sum\limits_{a,b,c} ab\\ a^2+b^2+c^2 &\geq ab+bc+ca\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca &\geq 0\\ \left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &\geq 0 \end{align*} $$

Which is obviously true.


However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:

$$ \begin{align*} \sum\limits_{a,b,c} a^3 &\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &\geq \sum\limits_{a,b,c} b\\ a+b+c &\geq a+b+c \end{align*} $$

Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?

Edit: Please don't help me with the original inequality, I'll figure it out.

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You can't assume what you want to prove. –  user60887 Jul 7 '13 at 23:26
    
@user60887 I'm not doing that, I'm trying to reduce it to something that I can prove. –  timvermeulen Jul 7 '13 at 23:27
    
@timvermeulen You cannot divide with $a$ the cyclic sum is an simpler way to write to expressiong $a^3+b^3+c^3$ since you cannot divide with $a$ in this expression you cannot divide in your other expression(with the cyclic sum symbol). Until you feel comfortable with another way of writing the same thing, first translate what an operation means in the expression where you are familiar with. –  clark Jul 7 '13 at 23:31
    
for example write $\sum _{a,b,c} a^2 \geq \sum_ {a,b,c}a, \Rightarrow \sum _{a,b,c} a \geq \sum_ {a,b,c}1$ which is false for $a=b=c=0$ –  clark Jul 7 '13 at 23:34
    
The inequality is obviously true if a=b=c so due to symmetry, why not consider a>b ? That is, write a = b + k with k>0 substitute for a and see if the inequality becomes easier to handle. (It is just a hunch, I am not sure if it works...) –  imranfat Jul 7 '13 at 23:41

3 Answers 3

up vote 5 down vote accepted

Just assume, wlog $a\leq b\leq c$. Then this equation is all you need: $$a^3+b^3+c^3=a^2b+b^2c+c^2a+\underset{\geq 0}{\underbrace{(c^2-a^2)(b-a)}}+\underset{\geq 0}{\underbrace{(c^2-b^2)(c-b)}}\geq a^2b+b^2c+c^2a$$

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but if $b \le a \le c$, this method doesn't work. –  chenbai Jul 9 '13 at 8:30
1  
This is what the wlog is about. As the equation is somehow symmetrical, you can use $$a^3+b^3+c^3=a^2b+b^2c+c^2a+(c^2-b^2)(c-a)+(a^2-b^2)(a-b)$$ in this case. –  Tomas Jul 9 '13 at 8:59
    
OK, that is nice. I simply swap $a$ and $b$ and get wrong result. –  chenbai Jul 9 '13 at 9:06
    
This is because, it is not totally symmetrical in $a^2b+b^2c+c^2a$, you do not get the same expression here by arbitrarily swapping. –  Tomas Jul 9 '13 at 9:16

Without making any assumption, just simple $AM\ge GM$ $$a^3+a^3+b^3\ge3a^2b$$ $$b^3+b^3+c^3\ge3b^2c$$ $$c^3+c^3+a^3\ge3c^2a$$ $$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$$

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WOLG, Let $c$=Max{$a,b,c$}, then there is 2 cases:

case I: $0<a \le b \le c$, we want to prove $c^2(c-a) \ge a^2(b-a)+b^2(c-b)$

we have $c^2\ge b^2, c^2\ge a^2 \to $,RHS $\le c^2(b-a)+c^2(c-b)=c^2(c-a)$

case II: $0<b \le a \le c$, we want to prove $a^2(a-b)+c^2(c-a) \ge b^2(c-b)$

we have $a^2\ge b^2,c^2 \ge b^2, \to$LHS $ \ge b^2(a-b)+b^2(c-a)=b^2(c-b)$

to summary 2 cases, we have $a^2(a-b)+b^2(b-c)+c^2(c-a) \ge 0$

QED

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