Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 8888 by 7C;00 last updated on 04/Nov/16

Commented by prakash jain last updated on 04/Nov/16

for positive numbers: Arithmetic mean ≥ Harmonic mean AM=((a_1 +a_2 +...+a_n )/n)=(n/n)=1 HM=(n/((1/a_1 )+(1/a_2 )+(1/a_3 )+...+(1/a_n ))) 1≥(n/((1/a_1 )+(1/a_2 )+(1/a_3 )+...+(1/a_n ))) ⇒(1/a_1 )+(1/a_2 )+(1/a_3 )+...+(1/a_n )≥n

$$\mathrm{for}\:\mathrm{positive}\:\mathrm{numbers}: \\ $$$$\mathrm{Arithmetic}\:\mathrm{mean}\:\geqslant\:\mathrm{Harmonic}\:\mathrm{mean} \\ $$$$\mathrm{AM}=\frac{{a}_{\mathrm{1}} +{a}_{\mathrm{2}} +...+{a}_{{n}} }{{n}}=\frac{{n}}{{n}}=\mathrm{1} \\ $$$$\mathrm{HM}=\frac{{n}}{\frac{\mathrm{1}}{{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} }+...+\frac{\mathrm{1}}{{a}_{{n}} }} \\ $$$$\mathrm{1}\geqslant\frac{{n}}{\frac{\mathrm{1}}{{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} }+...+\frac{\mathrm{1}}{{a}_{{n}} }} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}_{\mathrm{1}} }+\frac{\mathrm{1}}{{a}_{\mathrm{2}} }+\frac{\mathrm{1}}{{a}_{\mathrm{3}} }+...+\frac{\mathrm{1}}{{a}_{{n}} }\geqslant{n} \\ $$

Answered by Chantria last updated on 08/Nov/16

Solution. for a_1 ,a_2 ,...,a_n >0 and b_1 ,b_2 ,...,b_n >0 by Cauchy−Schwarz in eagle form (a_1 ^2 /b_1 )+(a_2 ^2 /b_2 )+...+(a_n ^2 /b_n )≥(((a_1 +a_2 +...+a_n )^2 )/(b_1 +b_2 +...+b_n )) (1^2 /a_1 )+(1^2 /a_2 )+...+(1^2 /a_n )≥(((1+1+...+1)^2 )/(a_1 +a_2 +...+a_n ))=(n^2 /n)=n because a_1 +a_2 +...+a_n =n

$$\boldsymbol{\mathrm{Solution}}. \\ $$$$\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{a}}_{\mathrm{1}} ,\boldsymbol{\mathrm{a}}_{\mathrm{2}} ,...,\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} >\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{b}}_{\mathrm{1}} ,\boldsymbol{\mathrm{b}}_{\mathrm{2}} ,...,\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} >\mathrm{0} \\ $$$$\boldsymbol{\mathrm{by}}\:\boldsymbol{\mathrm{Cauchy}}−\boldsymbol{\mathrm{Schwarz}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{eagle}}\:\boldsymbol{\mathrm{form}} \\ $$$$\:\frac{\boldsymbol{\mathrm{a}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}_{\mathrm{1}} }+\frac{\boldsymbol{\mathrm{a}}_{\mathrm{2}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}_{\mathrm{2}} }+...+\frac{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} }\geqslant\frac{\left(\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +...+\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}_{\mathrm{1}} +\boldsymbol{\mathrm{b}}_{\mathrm{2}} +...+\boldsymbol{\mathrm{b}}_{\boldsymbol{\mathrm{n}}} } \\ $$$$\frac{\mathrm{1}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}_{\mathrm{1}} }+\frac{\mathrm{1}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}_{\mathrm{2}} }+...+\frac{\mathrm{1}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} }\geqslant\frac{\left(\mathrm{1}+\mathrm{1}+...+\mathrm{1}\right)^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +...+\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} }=\frac{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }{\boldsymbol{\mathrm{n}}}=\boldsymbol{\mathrm{n}} \\ $$$$\:\boldsymbol{\mathrm{because}}\:\boldsymbol{\mathrm{a}}_{\mathrm{1}} +\boldsymbol{\mathrm{a}}_{\mathrm{2}} +...+\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{n}}} =\boldsymbol{\mathrm{n}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com