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Question-78049

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Question Number 78049 by Pratah last updated on 13/Jan/20
Commented by Pratah last updated on 14/Jan/20
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Answered by behi83417@gmail.com last updated on 14/Jan/20
for:n=1⇒(1/(1/(1+a_1 )))−(1/a_1 )=1+a_1 −(1/a_1 )≥1(a_1 ∈N) for:n=2⇒(1/(1/(1+a_1 )))+(1/(1/(1+a_1 )))−(1/((1/a_1 )+(1/a_2 )))= =2+a_1 +a_2 −((a_1 .a_2 )/(a_1 +a_2 ))≥2+a_1 +a_2 −(1/2)>(1/2)(a_i ∈N) it is true for:n=1,2.let assume that is true for:n=k∈N.now we show that it is true for:n=k+1 n=k+1⇒(1/(1/(1+a_1 )))+(1/(1/(1+a_2 )))+..+(1/(1/(1+a_(k+1) )))− −(1/((1/a_1 )+(1/a_2 )+(1/a_3 )+.....+(1/a_(k+1) )))≥(1/n)+1+a_(k+1) −(1/(n+1)) =1+a_(k+1) +(1/(n(n+1)))>(1/(n+1)) .■
$$\mathrm{for}:\mathrm{n}=\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{1}} }}−\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{1}} }=\mathrm{1}+\mathrm{a}_{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{1}} }\geqslant\mathrm{1}\left(\mathrm{a}_{\mathrm{1}} \in\mathrm{N}\right) \\ $$$$\mathrm{for}:\mathrm{n}=\mathrm{2}\Rightarrow\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{1}} }}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{1}} }}−\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{2}} }}= \\ $$$$=\mathrm{2}+\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} −\frac{\mathrm{a}_{\mathrm{1}} .\mathrm{a}_{\mathrm{2}} }{\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} }\geqslant\mathrm{2}+\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}>\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}_{\mathrm{i}} \in\mathrm{N}\right) \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}:\mathrm{n}=\mathrm{1},\mathrm{2}.\mathrm{let}\:\mathrm{assume}\:\mathrm{that}\:\mathrm{is} \\ $$$$\mathrm{true}\:\mathrm{for}:\mathrm{n}=\mathrm{k}\in\mathrm{N}.\mathrm{now}\:\mathrm{we}\:\mathrm{show}\:\mathrm{that}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{true}\:\mathrm{for}:\mathrm{n}=\mathrm{k}+\mathrm{1} \\ $$$$\mathrm{n}=\mathrm{k}+\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{1}} }}+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{2}} }}+..+\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{a}_{\mathrm{k}+\mathrm{1}} }}− \\ $$$$−\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{3}} }+…..+\frac{\mathrm{1}}{\mathrm{a}_{\mathrm{k}+\mathrm{1}} }}\geqslant\frac{\mathrm{1}}{\mathrm{n}}+\mathrm{1}+\mathrm{a}_{\mathrm{k}+\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}} \\ $$$$=\mathrm{1}+\mathrm{a}_{\mathrm{k}+\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)}>\frac{\mathrm{1}}{\mathrm{n}+\mathrm{1}}\:\:\:\:.\blacksquare \\ $$
Commented by mr W last updated on 14/Jan/20
great! i think the inequality is even valid, if a_i are not positive intergers, but any positive real numbers.
$${great}! \\ $$$${i}\:{think}\:{the}\:{inequality}\:{is}\:{even}\:{valid}, \\ $$$${if}\:{a}_{{i}} \:{are}\:{not}\:{positive}\:{intergers},\:{but} \\ $$$${any}\:{positive}\:{real}\:{numbers}. \\ $$
Answered by mr W last updated on 14/Jan/20
f(a_i )=(1/(Σ(1/(1+a_i ))))−(1/(Σ(1/a_i ))) (∂f/∂a_i )=(1/((Σ(1/(1+a_i )))^2 ))×((−1)/((1+a_i )^2 ))−(1/((Σ(1/a_i ))^2 ))×((−1)/a_i ^2 )=0 ⇒((Σ(1/a_i ))/(Σ(1/(1+a_i ))))=((1+a_i )/a_(i ) ) ⇒(1/a_i )=((Σ(1/a_i ))/(Σ(1/(1+a_i ))))−1 i.e. for minimum of f(a_i ) all a_i should be equal: a_1 =a_2 =...=a_n =a>0 min.f=(1/(n/(1+a)))−(1/(n/a))=((1+a)/n)−(a/n)=(1/n) ⇒f(a_i )≥min.f=(1/n)
$${f}\left({a}_{{i}} \right)=\frac{\mathrm{1}}{\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}_{{i}} }}−\frac{\mathrm{1}}{\Sigma\frac{\mathrm{1}}{{a}_{{i}} }} \\ $$$$\frac{\partial{f}}{\partial{a}_{{i}} }=\frac{\mathrm{1}}{\left(\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}_{{i}} }\right)^{\mathrm{2}} }×\frac{−\mathrm{1}}{\left(\mathrm{1}+{a}_{{i}} \right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\left(\Sigma\frac{\mathrm{1}}{{a}_{{i}} }\right)^{\mathrm{2}} }×\frac{−\mathrm{1}}{{a}_{{i}} ^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\Sigma\frac{\mathrm{1}}{{a}_{{i}} }}{\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}_{{i}} }}=\frac{\mathrm{1}+{a}_{{i}} }{{a}_{{i}\:} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}_{{i}} }=\frac{\Sigma\frac{\mathrm{1}}{{a}_{{i}} }}{\Sigma\frac{\mathrm{1}}{\mathrm{1}+{a}_{{i}} }}−\mathrm{1} \\ $$$${i}.{e}.\:{for}\:{minimum}\:{of}\:{f}\left({a}_{{i}} \right)\:{all}\:{a}_{{i}} \:{should} \\ $$$${be}\:{equal}:\:{a}_{\mathrm{1}} ={a}_{\mathrm{2}} =…={a}_{{n}} ={a}>\mathrm{0} \\ $$$${min}.{f}=\frac{\mathrm{1}}{\frac{{n}}{\mathrm{1}+{a}}}−\frac{\mathrm{1}}{\frac{{n}}{{a}}}=\frac{\mathrm{1}+{a}}{{n}}−\frac{{a}}{{n}}=\frac{\mathrm{1}}{{n}} \\ $$$$\Rightarrow{f}\left({a}_{{i}} \right)\geqslant{min}.{f}=\frac{\mathrm{1}}{{n}} \\ $$
Commented by mr W last updated on 14/Jan/20
for this proof we only need a_i ∈R^+ instead of a_i ∈N.
$${for}\:{this}\:{proof}\:{we}\:{only}\:{need}\:{a}_{{i}} \in{R}^{+} \\ $$$${instead}\:{of}\:{a}_{{i}} \in\mathbb{N}. \\ $$

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