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TR-MO 2008 Problem 3

    a+b+c=1a+b+c=1 koşulunu sağlayan tüm aa, bb, cc, pozitif gerçel sayıları için,

    a2b2c3(a2−ab+b2)+b2c2a3(b2−bc+c2)+c2a2b3(c2−ca+a2)≥3ab+bc+ca\frac{a^2b^2}{c^3(a^2-ab+b^2)}+\frac{b^2c^2}{a^3(b^2-bc+c^2)}+\frac{c^2a^2}{b^3(c^2-ca+a^2)} \geq \frac{3}{ab+bc+ca}

    olduğunu kanıtlayınız.

    Official text Turkish Mathematical Olympiad 2008 official paper, page 1. Quoted with source.

    A model read this text from the scan of the paper by OCR, with vision/claude@2.1.318 (claude-opus-5-5). No person has checked it against the paper yet.

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