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let-a-b-c-gt-0-a-b-c-3-prove-that-a-2-b-2-bc-c-2-b-2-c-2-ac-a-2-c-2-b-2-ba-a-2-6-2abc-5-3-a-2-b-2-c-2-7-Z-A-




Question Number 164405 by Zaynal last updated on 16/Jan/22
let a, b, c > 0 ; a + b + c = 3  prove that; (a^2 /(b^2  + bc + c^2 ))  +  (b^2 /(c^2  + ac + a^2  ))  +  (c^2 /(b^2  +ba + a^2 ))  6  ≥  2abc  +  (5/3)  (a^2  + b^2  + c^2  )  ≥  7  ^({Z.A})
$$\boldsymbol{{let}}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:>\:\mathrm{0}\:;\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{c}}\:=\:\mathrm{3} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}};\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} \:+\:\boldsymbol{{bc}}\:+\:\boldsymbol{{c}}^{\mathrm{2}} }\:\:+\:\:\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{c}}^{\mathrm{2}} \:+\:\boldsymbol{{ac}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} \:}\:\:+\:\:\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} \:+\boldsymbol{{ba}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} }\:\:\mathrm{6}\:\:\geqslant\:\:\mathrm{2}\boldsymbol{{abc}}\:\:+\:\:\frac{\mathrm{5}}{\mathrm{3}}\:\:\left(\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\boldsymbol{{c}}^{\mathrm{2}} \:\right)\:\:\geqslant\:\:\mathrm{7} \\ $$$$\:^{\left\{\boldsymbol{\mathrm{Z}}.\mathrm{A}\right\}} \\ $$

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