3
$\begingroup$

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.

$\endgroup$
8
  • 2
    $\begingroup$ You can't assume what you want to prove. $\endgroup$
    – user60887
    Commented Jul 7, 2013 at 23:26
  • $\begingroup$ @user60887 I'm not doing that, I'm trying to reduce it to something that I can prove. $\endgroup$ Commented Jul 7, 2013 at 23:27
  • 1
    $\begingroup$ @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. $\endgroup$
    – clark
    Commented Jul 7, 2013 at 23:31
  • $\begingroup$ 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 $a=b=c=0$ $\endgroup$
    – clark
    Commented Jul 7, 2013 at 23:34
  • $\begingroup$ 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...) $\endgroup$
    – imranfat
    Commented Jul 7, 2013 at 23:41

4 Answers 4

23
$\begingroup$

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

$\endgroup$
6
$\begingroup$

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

$\endgroup$
4
  • 1
    $\begingroup$ but if $b \le a \le c$, this method doesn't work. $\endgroup$
    – chenbai
    Commented Jul 9, 2013 at 8:30
  • 2
    $\begingroup$ 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. $\endgroup$
    – Tomas
    Commented Jul 9, 2013 at 8:59
  • $\begingroup$ OK, that is nice. I simply swap $a$ and $b$ and get wrong result. $\endgroup$
    – chenbai
    Commented Jul 9, 2013 at 9:06
  • $\begingroup$ 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. $\endgroup$
    – Tomas
    Commented Jul 9, 2013 at 9:16
1
$\begingroup$

(@HaiDangel told me. https://diendantoanhoc.net/topic/182934-a3-b3-c3geqq-a2b-b2c-c2a/?p=731023)

A stronger version: Let $a, b, c$ be real numbers with $a + b \ge 0, b + c \ge 0$ and $c+a\ge 0$. Prove that $$a^3 + b^3 + c^3 \ge a^2b + b^2c + c^2a.$$

I have an SOS expression: \begin{align} &a^3 + b^3 + c^3 - a^2b - b^2c - c^2a \\ =\ & \frac{(a^2+b^2-2c^2)^2 + 3(a^2-b^2)^2 + \sum_{\mathrm{cyc}} 4(a+b)(c+a)(a-b)^2}{8(a+b+c)}. \end{align}

$\endgroup$
0
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .