Prove that the positive numbers $\displaystyle a, b $ and $\displaystyle c$ satisfy the inequality $\displaystyle \frac{1}{a^{3}+b^{3}+abc }+\frac{1}{b^{3}+c^{3}+abc}+\frac{1}{c^{3}+a^{3}+a bc}\leq\frac{1}{abc}$
Follow Math Help Forum on Facebook and Google+
$\displaystyle x^3 + y^3 \ge xy(x + y)$ Therefore $\displaystyle \sum \frac {1}{a^3 + b^3 + abc} \leq \sum \frac {1}{ab(a + b + c)} = \sum \frac {c}{abc(a + b + c)} = \frac {1}{abc}$
can you tell me a little bit more why it is so please
Which step didn't you understand, I'll go into that in more detail
i didn't understand this step : Originally Posted by SimonM $\displaystyle \sum \frac {1}{a^3 + b^3 + abc} \leq \sum \frac {1}{ab(a + b + c)} = \sum \frac {c}{abc(a + b + c)} = \frac {1}{abc}$
$\displaystyle a^3+b^3+abc \ge ab(a+b)+abc = ab(a+b+c)$ Now take the reciprocal
and why this : $\displaystyle \sum \frac{c}{abc(a + b + c)} = \frac {1}{abc} $
$\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)} $
View Tag Cloud