1. ## Proof q.

a, b, c are three positive numbers, no two of which are equal. Prove that
a^2 - ab + b^2 > ab
a^3 + b^3 > ab (a + b)
2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)

2. Originally Posted by Detanon
a, b, c are three positive numbers, no two of which are equal. Prove that
a^2 - ab + b^2 > ab
This is equivalent to $\displaystyle (a-b)^2 > 0$
a^3 + b^3 > ab (a + b)
Multiply the preceding inequality by $\displaystyle a+b$
2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)
From the preceding,
$\displaystyle a^3 + b^3 > ab \; (a + b)$
$\displaystyle a^3 + c^3 > ac \; (a + c)$
$\displaystyle b^3 + c^3 > bc \; (b + c)$
(Just substituting different variables)

3. Why is $\displaystyle (a-b)^2 > 0$ ??
Is it because its a perfect square or something so its greater than zero?

I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)

4. Originally Posted by Detanon
Why is $\displaystyle (a-b)^2 > 0$ ??
Is it because its a perfect square or something so its greater than zero?
The square of any non-zero number is positive. We know $\displaystyle a - b \neq 0$ because we were told $\displaystyle a \neq b$.

I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)
Multiply $\displaystyle a^2 -ab +b^2$ by $\displaystyle a+b$. What do you get?

5. a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please.

6. Originally Posted by Detanon
a^3 + b^3? But why do you multiply a^2 -ab +b^2 by a+b. Can you show all the steps please.
I have listed all the steps.

I suggest you go back and re-read my first post.