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

a,b,c,d,e,f, are + real numbers prove: (a/(b+c))+(b/(c+d))+(c/(d+e))+(d/(e+f))+(e/(f+a))+(f/(a+b))≥3

$${a},{b},{c},{d},{e},{f},\:{are}\:+\:{real}\:{numbers} \\ $$$${prove}: \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{d}}+\frac{{c}}{{d}+{e}}+\frac{{d}}{{e}+{f}}+\frac{{e}}{{f}+{a}}+\frac{{f}}{{a}+{b}}\geqslant\mathrm{3} \\ $$

Answered by deleteduser1 last updated on 23/Jun/23

=Σ_(cyc) (a^2 /(a(b+c))) ≥(((Σa)^2 )/(Σ_(cyc) a(b+c)))....(i) 2Σ_(cyc) a(b+c)=(Σa)^2 −(a+d)^2 −(b+e)^2 −(c+f)^2 From 3(a^2 +b^2 +c^2 )≥(a+b+c)^2 We get 2Σ_(cyc) a(b+c)≤(2/3)(Σa)^2 ...(ii) (i) and (ii)⇒Σ(a^2 /(a(b+c)))≥((2(Σa)^2 )/(2Σa(b+c)))≥3 □

$$=\underset{{cyc}} {\sum}\frac{{a}^{\mathrm{2}} }{{a}\left({b}+{c}\right)}\:\geqslant\frac{\left(\Sigma{a}\right)^{\mathrm{2}} }{\underset{{cyc}} {\sum}{a}\left({b}+{c}\right)}....\left({i}\right) \\ $$$$\mathrm{2}\underset{{cyc}} {\sum}{a}\left({b}+{c}\right)=\left(\Sigma{a}\right)^{\mathrm{2}} −\left({a}+{d}\right)^{\mathrm{2}} −\left({b}+{e}\right)^{\mathrm{2}} −\left({c}+{f}\right)^{\mathrm{2}} \\ $$$${From}\:\mathrm{3}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\geqslant\left({a}+{b}+{c}\right)^{\mathrm{2}} \\ $$$${We}\:{get}\:\mathrm{2}\underset{{cyc}} {\sum}{a}\left({b}+{c}\right)\leqslant\frac{\mathrm{2}}{\mathrm{3}}\left(\Sigma{a}\right)^{\mathrm{2}} ...\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\Rightarrow\Sigma\frac{{a}^{\mathrm{2}} }{{a}\left({b}+{c}\right)}\geqslant\frac{\mathrm{2}\left(\Sigma{a}\right)^{\mathrm{2}} }{\mathrm{2}\Sigma{a}\left({b}+{c}\right)}\geqslant\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Box \\ $$

Commented by Subhi last updated on 23/Jun/23

perfect

$${perfect}\: \\ $$

Answered by Subhi last updated on 23/Jun/23

Σ_(cyc) a(b+c)Σ_(cyc) (((a))/((b+c)))≥(Σa)^2 Σ(((a))/((b+c)))≥(((Σa)^2 )/(Σa(b+c)))=(((a+b+c+d+e+f)^2 )/(ab+ac+bc+bd+cd+ce+de+df+ef+ea+af+fb)) 2Σa(b+c)=(Σ_(cyc) a)^2 −Σa^2 −2(ad+be+cf)=(Σ_(cyc) a)^2 −((a+d)^2 +(b+e)^2 +(c+f)^2 ) (a+d)^2 +(b+e)^2 +(c+f)^2 ≥(((a+b+c+d+e+f)^2 )/3) 2Σa(b+c)≤(2/3)(Σa)^2 ⇛ Σa(b+c)≤(1/3)(Σa)^2 ∴ Σ_(cyc) (a/(b+c))≥((3(Σa)^2 )/((Σa)^2 ))=3

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

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