Question Number 196870 by York12 last updated on 02/Sep/23
$${let}\:{b}_{{i}} \wedge\:{a}_{{i}} >\mathrm{0}\:{where}\:{i}\in\left\{\mathrm{1},\mathrm{2},\mathrm{3},…,{n}\right\}\&\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({b}_{{i}} \right)=\lambda\:{Prove}\:{that} \\ $$$$\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{2}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)+\frac{\lambda−\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{1}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)+….+\frac{\lambda−\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}{\left({b}_{\mathrm{2}} +{b}_{\mathrm{3}} \right)}\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)+…\frac{\lambda−\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}{\left({b}_{{n}−\mathrm{1}} +{b}_{{n}} \right)}\left({a}_{{n}−\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\geqslant\sqrt{\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}×\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left({a}_{{i}} {a}_{{j}} \right)} \\ $$$$ \\ $$
Answered by York12 last updated on 02/Sep/23
![WLOG: a_1 +a_2 =a_1 +a_3 =a_1 +a_4 =.....=1 ⇒Σ_(i=1) ^n (a_i )=(n/2) (√((Σ_i ^n (a_i ^2 ))/n))≥((Σ_(i=1) ^n (a_i ))/n) ⇒Σ_i ^n (a_i ^2 )≥(([Σ_(i=1) ^n (a_i )]^2 )/n) ⇒(((n−1)[Σ_(i=1) ^n (a_i )]^2 )/n)≥ΣΣ_(1≤i<j≤n) (a_i a_j )→.....(i) ((λ−(a_1 +a_2 ))/((a_1 +a_2 )))+((λ−(a_1 +a_3 ))/((a_1 +a_3 )))+...+((λ−(a_2 +a_3 ))/((a_2 +a_3 )))+...+(((n(n−1))/2))−(((n(n−1))/2)) =λ((1/((a_1 +a_2 )))+(1/((a_1 +a_3 )))+...+(1/((a_2 +a_3 )))+...+(1/(a_(n−1) +a_n )))−(((n(n−1))/2)) ≥λ((((((n(n−1))/2))^2 )/(λ(n−1))))−(((n(n−1))/2))=((n(n−1)(n−2))/4)→....(ii) ((λ−(a_1 +a_2 ))/((a_1 +a_2 )))+((λ−(a_1 +a_3 ))/((a_1 +a_3 )))+...+((λ−(a_2 +a_3 ))/((a_2 +a_3 )))+...+((λ−(a_(n−1) +a_n ))/((a_(n−1) +a_n )))−(√(((n(n−1)(n−2)^2 )/4)ΣΣ_(1≤i<j≤n) (a_i a_j ))) ≥((n(n−1)(n−2))/4)−(√((n^2 (n−1)^2 (n−2)^2 )/(16))) ≥((n(n−1)(n−2))/4)−((n(n−1)(n−2))/4)=0 (Hence Proved)](https://www.tinkutara.com/question/Q196871.png)
$$ \\ $$$${WLOG}:\:{a}_{\mathrm{1}} +{a}_{\mathrm{2}} ={a}_{\mathrm{1}} +{a}_{\mathrm{3}} ={a}_{\mathrm{1}} +{a}_{\mathrm{4}} =…..=\mathrm{1} \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{i}} \right)=\frac{{n}}{\mathrm{2}} \\ $$$$\sqrt{\frac{\underset{{i}} {\overset{{n}} {\sum}}\left({a}_{{i}} ^{\mathrm{2}} \right)}{{n}}}\geqslant\frac{\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{i}} \right)}{{n}}\:\Rightarrow\underset{{i}} {\overset{{n}} {\sum}}\left({a}_{{i}} ^{\mathrm{2}} \right)\geqslant\frac{\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{i}} \right)\right]^{\mathrm{2}} }{{n}} \\ $$$$\Rightarrow\frac{\left({n}−\mathrm{1}\right)\left[\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\left({a}_{{i}} \right)\right]^{\mathrm{2}} }{{n}}\geqslant\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left({a}_{{i}} {a}_{{j}} \right)\rightarrow…..\left({i}\right) \\ $$$$\frac{\lambda−\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)}+\frac{\lambda−\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)}+…+\frac{\lambda−\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)}{\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)}+…+\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right)−\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$=\lambda\left(\frac{\mathrm{1}}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)}+\frac{\mathrm{1}}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)}+…+\frac{\mathrm{1}}{\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)}+…+\frac{\mathrm{1}}{{a}_{{n}−\mathrm{1}} +{a}_{{n}} }\right)−\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right) \\ $$$$\geqslant\lambda\left(\frac{\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }{\lambda\left({n}−\mathrm{1}\right)}\right)−\left(\frac{{n}\left({n}−\mathrm{1}\right)}{\mathrm{2}}\right)=\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{4}}\rightarrow….\left({ii}\right) \\ $$$$\frac{\lambda−\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} \right)}+\frac{\lambda−\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)}{\left({a}_{\mathrm{1}} +{a}_{\mathrm{3}} \right)}+…+\frac{\lambda−\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)}{\left({a}_{\mathrm{2}} +{a}_{\mathrm{3}} \right)}+…+\frac{\lambda−\left({a}_{{n}−\mathrm{1}} +{a}_{{n}} \right)}{\left({a}_{{n}−\mathrm{1}} +{a}_{{n}} \right)}−\sqrt{\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{4}}\underset{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {\Sigma\Sigma}\left({a}_{{i}} {a}_{{j}} \right)} \\ $$$$\geqslant\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{4}}−\sqrt{\frac{{n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \left({n}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{16}}} \\ $$$$\geqslant\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{4}}−\frac{{n}\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)}{\mathrm{4}}=\mathrm{0} \\ $$$$\left({Hence}\:{Proved}\right) \\ $$