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Question Number 193875 by Subhi last updated on 22/Jun/23

$$a,b,c,d,e,f,are+realnumbers$$$$prove:$$$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+e}+\frac{d}{e+f}+\frac{e}{f+a}+\frac{f}{a+b}⩾3$$

Answered by deleteduser1 last updated on 23/Jun/23

$$=\sum _{cyc}\frac{a^2}{a(b+c)}⩾\frac{{\left(\mathrm{\Sigma }a\right)}^2}{\sum _{cyc}a(b+c)}....\left(i\right)$$$$2\sum _{cyc}a(b+c)={\left(\mathrm{\Sigma }a\right)}^2-{(a+d)}^2-{(b+e)}^2-{(c+f)}^2$$$$From3(a^2+b^2+c^2)⩾{(a+b+c)}^2$$$$Weget2\sum _{cyc}a(b+c)⩽\frac{2}{3}{\left(\mathrm{\Sigma }a\right)}^2...\left(ii\right)$$$$\left(i\right)and\left(ii\right)\Rightarrow \mathrm{\Sigma }\frac{a^2}{a(b+c)}⩾\frac{2{\left(\mathrm{\Sigma }a\right)}^2}{2\mathrm{\Sigma }a(b+c)}⩾3◻$$

Commented by Subhi last updated on 23/Jun/23

$$perfect$$

Answered by Subhi last updated on 23/Jun/23

$$\sum _{cyc}a(b+c)\sum _{cyc}\frac{\left(a\right)}{(b+c)}⩾{\left(\mathrm{\Sigma }a\right)}^2$$$$\mathrm{\Sigma }\frac{\left(a\right)}{(b+c)}⩾\frac{{\left(\mathrm{\Sigma }a\right)}^2}{\mathrm{\Sigma }a(b+c)}=\frac{{(a+b+c+d+e+f)}^2}{ab+ac+bc+bd+cd+ce+de+df+ef+ea+af+fb}$$$$2\mathrm{\Sigma }a(b+c)={\left(\sum _{cyc}a\right)}^2-\mathrm{\Sigma }{a}^2-2(ad+be+cf)={\left(\sum _{cyc}a\right)}^2-({(a+d)}^2+{(b+e)}^2+{(c+f)}^2)$$$${(a+d)}^2+{(b+e)}^2+{(c+f)}^2⩾\frac{{(a+b+c+d+e+f)}^2}{3}$$$$2\mathrm{\Sigma }a(b+c)⩽\frac{2}{3}{\left(\mathrm{\Sigma }a\right)}^2\Rrightarrow \mathrm{\Sigma }a(b+c)⩽\frac{1}{3}{\left(\mathrm{\Sigma }a\right)}^2$$$$\therefore \sum _{cyc}\frac{a}{b+c}⩾\frac{3{\left(\mathrm{\Sigma }a\right)}^2}{{\left(\mathrm{\Sigma }a\right)}^2}=3$$

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