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}

Collected in the board: Inequality

Steven Zheng posted 1 month ago


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}

Steven Zheng posted 1 month ago

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