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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 =?

$${for}\:{a},{b},{c},{d},{e}\:\in{R}\:{and}\:{a}+{b}+{c}+{d}+{e}=\mathrm{5} \\ $$$${find}\:{the}\:{minimum}\:{value}\:{of}\: \\ $$$${a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} +\mathrm{5}{e}^{\mathrm{2}} =? \\ $$

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))

$$\mathrm{a}^{\mathrm{2}} +\mathrm{2b}^{\mathrm{2}} +\mathrm{3c}^{\mathrm{2}} +\mathrm{4d}^{\mathrm{2}} +\mathrm{5e}^{\mathrm{2}} \\ $$$$=\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{b}^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{2}}}+\frac{\mathrm{c}^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{3}}}+\frac{\mathrm{c}^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{4}}}+\frac{\mathrm{e}^{\mathrm{2}} }{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$$\geqslant\frac{\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}\right)^{\mathrm{2}} }{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$$=\frac{\mathrm{1500}}{\mathrm{137}} \\ $$$$\mathrm{inequality}\:\mathrm{hold}\:\mathrm{only}\:\frac{\mathrm{a}}{\mathrm{1}}=\frac{\mathrm{b}}{\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{c}}{\frac{\mathrm{1}}{\mathrm{3}}}=\frac{\mathrm{d}}{\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\mathrm{e}}{\frac{\mathrm{1}}{\mathrm{5}}} \\ $$$$\mathrm{ie}.\:\mathrm{a}=\frac{\mathrm{300}}{\mathrm{137}}\:\:\:\mathrm{b}=\frac{\mathrm{150}}{\mathrm{137}}\:\:\:\mathrm{c}=\frac{\mathrm{100}}{\mathrm{137}}\:\:\:\mathrm{d}=\frac{\mathrm{75}}{\mathrm{137}}\:\:\mathrm{e}=\frac{\mathrm{60}}{\mathrm{137}} \\ $$

Commented by mr W last updated on 02/Oct/21

thanks sir!

$${thanks}\:{sir}! \\ $$

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))

$${using}\:{method}\:{as}\:{in}\:{Q}\mathrm{155495}: \\ $$$$ \\ $$$${S}={a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{3}{c}^{\mathrm{2}} +\mathrm{4}{d}^{\mathrm{2}} +\mathrm{5}{e}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} +\left(\frac{{b}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\right)^{\mathrm{2}} +\left(\frac{{c}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right)^{\mathrm{2}} +\left(\frac{{d}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}}\right)^{\mathrm{2}} +\left(\frac{{e}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}}\right)^{\mathrm{2}} \\ $$$$=\left({a}'\right)^{\mathrm{2}} +\left({b}'\right)^{\mathrm{2}} +\left({c}'\right)^{\mathrm{2}} +\left({d}'\right)^{\mathrm{2}} +\left({e}'\right)^{\mathrm{2}} \\ $$$$ \\ $$$${a}+{b}+{c}+{d}+{e}=\mathrm{5} \\ $$$${a}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\left(\frac{{b}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}×\left(\frac{{c}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}×\left(\frac{{d}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}×\left(\frac{{e}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}}\right)=\mathrm{5} \\ $$$$\Rightarrow{a}'+\frac{{b}'}{\:\sqrt{\mathrm{2}}}+\frac{{c}'}{\:\sqrt{\mathrm{3}}}+\frac{{d}'}{\:\sqrt{\mathrm{4}}}+\frac{{e}'}{\:\sqrt{\mathrm{5}}}=\mathrm{5} \\ $$$$ \\ $$$${S}_{{min}} =\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}\right)^{\mathrm{2}} }}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{5}}}=\frac{\mathrm{1500}}{\mathrm{137}} \\ $$

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