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Question Number 205643 by hardmath last updated on 26/Mar/24

$$\mathrm{If}\mathrm{a},\mathrm{b},\mathrm{c}>0\mathrm{and}{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2=\mathrm{abc}$$$$\mathrm{Prove}\mathrm{that}:$$$$\frac{\mathrm{a}}{{\mathrm{a}}^2+\mathrm{bc}}+\frac{\mathrm{b}}{{\mathrm{b}}^2+\mathrm{ac}}+\frac{\mathrm{c}}{{\mathrm{c}}^2+\mathrm{ab}}⩽\frac{1}{2}$$

Answered by A5T last updated on 29/Mar/24

$$\mathrm{\Sigma }\frac{a}{a^2+bc}⩽\sqrt{a^2+b^2+c^2}\sqrt{\mathrm{\Sigma }\frac{1}{{(a^2+bc)}^2}}$$$$⩽\sqrt{abc}\times \frac{1}{2\sqrt{abc}}\left(\sqrt{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\right)⩽\frac{1}{2}\left(\sqrt{\frac{ab+bc+ca}{abc}}\right)$$$$=\frac{1}{2}\sqrt{\frac{ab+bc+ca}{a^2+b^2+c^2}}⩽\frac{1}{2}$$

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