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Question Number 155568 by mr W last updated on 02/Oct/21
for a,b,c,d,e ∈R and a+b+c+d+e=5  find the minimum value of   a^2 +2b^2 +3c^2 +4d^2 +5e^2 =?
fora,b,c,d,eRanda+b+c+d+e=5findtheminimumvalueofa2+2b2+3c2+4d2+5e2=?
Answered by qaz last updated on 02/Oct/21
a^2 +2b^2 +3c^2 +4d^2 +5e^2   =a^2 +(b^2 /(1/2))+(c^2 /(1/3))+(c^2 /(1/4))+(e^2 /(1/5))  ≥(((a+b+c+d+e)^2 )/(1+(1/2)+(1/3)+(1/4)+(1/5)))  =((1500)/(137))  inequality hold only (a/1)=(b/(1/2))=(c/(1/3))=(d/(1/4))=(e/(1/5))  ie. a=((300)/(137))   b=((150)/(137))   c=((100)/(137))   d=((75)/(137))  e=((60)/(137))
a2+2b2+3c2+4d2+5e2=a2+b212+c213+c214+e215(a+b+c+d+e)21+12+13+14+15=1500137inequalityholdonlya1=b12=c13=d14=e15ie.a=300137b=150137c=100137d=75137e=60137
Commented by mr W last updated on 02/Oct/21
thanks sir!
thankssir!
Answered by mr W last updated on 02/Oct/21
using method as in Q155495:    S=a^2 +2b^2 +3c^2 +4d^2 +5e^2   =a^2 +((b/(1/( (√2)))))^2 +((c/(1/( (√3)))))^2 +((d/(1/( (√4)))))^2 +((e/(1/( (√5)))))^2   =(a′)^2 +(b′)^2 +(c′)^2 +(d′)^2 +(e′)^2     a+b+c+d+e=5  a+(1/( (√2)))×((b/(1/( (√2)))))+(1/( (√3)))×((c/(1/( (√3)))))+(1/( (√4)))×((d/(1/( (√4)))))+(1/( (√5)))×((e/(1/( (√5)))))=5  ⇒a′+((b′)/( (√2)))+((c′)/( (√3)))+(d′/( (√4)))+((e′)/( (√5)))=5    S_(min) =((5/( (√(1^2 +((1/( (√2))))^2 +((1/( (√3))))^2 +((1/( (√4))))^2 +((1/( (√5))))^2 )))))^2   =((25)/(1+(1/2)+(1/3)+(1/4)+(1/5)))=((1500)/(137))
usingmethodasinQ155495:S=a2+2b2+3c2+4d2+5e2=a2+(b12)2+(c13)2+(d14)2+(e15)2=(a)2+(b)2+(c)2+(d)2+(e)2a+b+c+d+e=5a+12×(b12)+13×(c13)+14×(d14)+15×(e15)=5a+b2+c3+d4+e5=5Smin=(512+(12)2+(13)2+(14)2+(15)2)2=251+12+13+14+15=1500137

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