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Question Number 152912 by mathdanisur last updated on 03/Sep/21

Answered by ghimisi last updated on 03/Sep/21

$$\mathrm{\Sigma }\frac{a}{2b+3c}=\mathrm{\Sigma }\frac{a^2}{2ab+3ac}⩾\frac{{(a+b+c)}^2}{5(ab+bc+ac)}⩾\frac{3(ab+bc+ca)}{5(ab+bc+ac)}=\frac{3}{5}$$

Commented by ghimisi last updated on 03/Sep/21

$$\mathrm{\Sigma }\frac{a}{3b+2c}=\mathrm{\Sigma }\frac{a^2}{3ab+2ac}⩾\frac{{(a+b+c)}^2}{5(ab+bc+ac)}⩾\frac{3(ab+bc+ca)}{5(ab+bc+ac)}=\frac{3}{5}$$$$$$

Commented by mathdanisur last updated on 03/Sep/21

$$\mathrm{Nice}\mathbf{S}\mathrm{er}\mathrm{thank}\mathrm{you}$$

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