Question
For any real numbers a, b, c, prove that
\dfrac{1}{a^3+b^3+abc}+ \dfrac{1}{b^3+c^3+abc} +\dfrac{1}{a^3+c^3+abc}\leq \dfrac{1}{abc}
Answer
Using Sum of Perfect Cubes formula
a^3+b^3=(a+b)(a^2-ab+b^2)
\geq (a+b)ab
\dfrac{1}{a^3+b^3+abc}\leq \dfrac{1}{ab(a+b)+abc} = \dfrac{1}{ab(a+b+c)}
Similarly,
\dfrac{1}{b^3+c^3+abc}\leq \dfrac{1}{bc(a+b+c)}
\dfrac{1}{a^3+c^3+abc}\leq \dfrac{1}{ac(a+b+c)}
Therefore,
\dfrac{1}{a^3+b^3+abc}+ \dfrac{1}{b^3+c^3+abc} + \dfrac{1}{a^3+c^3+abc}
\leq \dfrac{1}{ab(a+b+c)}+\dfrac{1}{bc(a+b+c)}+\dfrac{1}{ac(a+b+c)}
=\dfrac{1}{a+b+c}(\dfrac{1}{ab} +\dfrac{1}{bc}+\dfrac{1}{ac}) =\dfrac{1}{abc}