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)
This is equivalent to $\displaystyle (a-b)^2 > 0$
Multiply the preceding inequality by $\displaystyle a+b$a^3 + b^3 > ab (a + b)
From the preceding,2(a^3 + b^3 + c^3) > ab (a + b) + bc (b + c) + ca (c + a)
$\displaystyle a^3 + b^3 > ab \; (a + b)$
$\displaystyle a^3 + c^3 > ac \; (a + c)$
$\displaystyle b^3 + c^3 > bc \; (b + c)$
(Just substituting different variables)
Now add.
The square of any non-zero number is positive. We know $\displaystyle a - b \neq 0$ because we were told $\displaystyle a \neq b$.
Multiply $\displaystyle a^2 -ab +b^2$ by $\displaystyle a+b$. What do you get?I dont really get the second steps either.. what did you do with a^3 + b^3 > ab (a + b)