1. ## inequality

Prove that the positive numbers $\displaystyle a, b$ and $\displaystyle c$ satisfy the inequality

$\displaystyle \frac{1}{a^{3}+b^{3}+abc }+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+a bc}\leq\frac{1}{abc}$

2. $\displaystyle x^3 + y^3 \ge xy(x + y)$

Therefore

$\displaystyle \sum \frac {1}{a^3 + b^3 + abc} \leq \sum \frac {1}{ab(a + b + c)} = \sum \frac {c}{abc(a + b + c)} = \frac {1}{abc}$

3. can you tell me a little bit more why it is so please

4. Which step didn't you understand, I'll go into that in more detail

5. i didn't understand this step :
Originally Posted by SimonM

$\displaystyle \sum \frac {1}{a^3 + b^3 + abc} \leq \sum \frac {1}{ab(a + b + c)} = \sum \frac {c}{abc(a + b + c)} = \frac {1}{abc}$

6. $\displaystyle a^3+b^3+abc \ge ab(a+b)+abc = ab(a+b+c)$

Now take the reciprocal

7. and why this : $\displaystyle \sum \frac{c}{abc(a + b + c)} = \frac {1}{abc}$

8. $\displaystyle \sum \frac{c}{abc(a+b+c)} = \frac{c}{abc(a+b+c)} + \frac{a}{abc(a+b+c)} + \frac{b}{abc(a+b+c)} =\frac{c+a+b}{abc(a+b+c)}$