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Question-152912




Question Number 152912 by mathdanisur last updated on 03/Sep/21
Answered by ghimisi last updated on 03/Sep/21
Σ(a/(2b+3c))=Σ(a^2 /(2ab+3ac))≥(((a+b+c)^2 )/(5(ab+bc+ac)))≥((3(ab+bc+ca))/(5(ab+bc+ac)))=(3/5)
$$\Sigma\frac{{a}}{\mathrm{2}{b}+\mathrm{3}{c}}=\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{2}{ab}+\mathrm{3}{ac}}\geqslant\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{5}\left({ab}+{bc}+{ac}\right)}\geqslant\frac{\mathrm{3}\left({ab}+{bc}+{ca}\right)}{\mathrm{5}\left({ab}+{bc}+{ac}\right)}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$
Commented by ghimisi last updated on 03/Sep/21
Σ(a/(3b+2c))=Σ(a^2 /(3ab+2ac))≥(((a+b+c)^2 )/(5(ab+bc+ac)))≥((3(ab+bc+ca))/(5(ab+bc+ac)))=(3/5)
$$\Sigma\frac{{a}}{\mathrm{3}{b}+\mathrm{2}{c}}=\Sigma\frac{{a}^{\mathrm{2}} }{\mathrm{3}{ab}+\mathrm{2}{ac}}\geqslant\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} }{\mathrm{5}\left({ab}+{bc}+{ac}\right)}\geqslant\frac{\mathrm{3}\left({ab}+{bc}+{ca}\right)}{\mathrm{5}\left({ab}+{bc}+{ac}\right)}=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$ \\ $$
Commented by mathdanisur last updated on 03/Sep/21
Nice Ser thank you
$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er}\:\mathrm{thank}\:\mathrm{you} \\ $$

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