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Question Number 205643 by hardmath last updated on 26/Mar/24
If  a,b,c>0  and  a^2  + b^2  + c^2  = abc  Prove that:  (a/(a^2  + bc)) + (b/(b^2  + ac)) + (c/(c^2  + ab)) ≤ (1/2)
$$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:=\:\mathrm{abc} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}}{\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{bc}}\:+\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{ac}}\:+\:\frac{\mathrm{c}}{\mathrm{c}^{\mathrm{2}} \:+\:\mathrm{ab}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by A5T last updated on 29/Mar/24
Σ(a/(a^2 +bc))≤(√(a^2 +b^2 +c^2 ))(√(Σ(1/((a^2 +bc)^2 ))))  ≤(√(abc))×(1/(2(√(abc))))((√((1/a)+(1/b)+(1/c))))≤(1/2)((√((ab+bc+ca)/(abc))))  =(1/2)(√((ab+bc+ca)/(a^2 +b^2 +c^2 )))≤(1/2)
$$\Sigma\frac{{a}}{{a}^{\mathrm{2}} +{bc}}\leqslant\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\sqrt{\Sigma\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +{bc}\right)^{\mathrm{2}} }} \\ $$$$\leqslant\sqrt{{abc}}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{abc}}}\left(\sqrt{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}\right)\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\frac{{ab}+{bc}+{ca}}{{abc}}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{{ab}+{bc}+{ca}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\leqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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