Question Number 219935 by fantastic last updated on 03/May/25
Answered by MrGaster last updated on 04/May/25
Commented by MrGaster last updated on 04/May/25
![(2):Let b_j =(√a_j )⇒a_j =b_j ^2 ,Σb_j ^2 =1 LHS=Σ(b_j ^2 /( (√(1−b_j ^2 ))))≥(((Σb_j )^2 )/(Σ(√(1−b_j ^2 )))) (Titu^, s lemma) Σ(√(1−b_j ^2 ))≤(√(nΣ(1−b_j ^2 )))=(√(n(n−1))) ⇒LHS≥(((Σb_j )^2 )/( (√(n(n−1)))))≥((Σb_j )/( (√(n−1)))) ∵(Σb_j )^2 ≤nΣb_j ^2 =n ⇒Σ(a_j /( (√(1−a_j ))))≥(1/( (√(n−1))))Σ(√a_j ) [Q.E.D]](https://www.tinkutara.com/question/Q219953.png)
$$\left(\mathrm{2}\right):\mathrm{Let}\:{b}_{{j}} =\sqrt{{a}_{{j}} }\Rightarrow{a}_{{j}} ={b}_{{j}} ^{\mathrm{2}} ,\Sigma{b}_{{j}} ^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{LHS}=\Sigma\frac{{b}_{{j}} ^{\mathrm{2}} }{\:\sqrt{\mathrm{1}−{b}_{{j}} ^{\mathrm{2}} }}\geq\frac{\left(\Sigma{b}_{{j}} \right)^{\mathrm{2}} }{\Sigma\sqrt{\mathrm{1}−{b}_{{j}} ^{\mathrm{2}} }}\:\left(\mathrm{Titu}^{,} {s}\:\mathrm{lemma}\right) \\ $$$$\Sigma\sqrt{\mathrm{1}−{b}_{{j}} ^{\mathrm{2}} }\leq\sqrt{{n}\Sigma\left(\mathrm{1}−{b}_{{j}} ^{\mathrm{2}} \right)}=\sqrt{{n}\left({n}−\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{LHS}\geq\frac{\left(\Sigma{b}_{{j}} \right)^{\mathrm{2}} }{\:\sqrt{{n}\left({n}−\mathrm{1}\right)}}\geq\frac{\Sigma{b}_{{j}} }{\:\sqrt{{n}−\mathrm{1}}}\:\because\left(\Sigma{b}_{{j}} \right)^{\mathrm{2}} \leq{n}\Sigma{b}_{{j}} ^{\mathrm{2}} ={n} \\ $$$$\Rightarrow\Sigma\frac{{a}_{{j}} }{\:\sqrt{\mathrm{1}−{a}_{{j}} }}\geq\frac{\mathrm{1}}{\:\sqrt{{n}−\mathrm{1}}}\Sigma\sqrt{{a}_{{j}} } \\ $$$$\left[{Q}.\mathrm{E}.\mathrm{D}\right] \\ $$$$ \\ $$