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Question Number 196198 by liuxinnan last updated on 19/Aug/23

a + b + c + d = 4

p r o v e t h a t :

\frac{a}{a^{3} + 8} + \frac{b}{b^{3} + 8} + \frac{c}{c^{3} + 8} + \frac{d}{d^{3} + 8} \leq \frac{4}{9}

Answered by AST last updated on 19/Aug/23

a^{3} + 8 = a^{3} + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 ⩾ 9 \sqrt[3]{a}

\Rightarrow Σ \frac{a}{a^{3} + 8} ⩽ \frac{1}{9} (Σ a^{\frac{2}{3}}) ⩽ \frac{1}{9} {(8 \times \frac{a + b + c + d}{4})}^{\frac{2}{3}} = \frac{4}{9}

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