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Question Number 155568 by mr W last updated on 02/Oct/21

f o r a, b, c, d, e \in R a n d a + b + c + d + e = 5

f i n d t h e m i n i m u m v a l u e o f

a^{2} + 2 b^{2} + 3 c^{2} + 4 d^{2} + 5 e^{2} = ?

Answered by qaz last updated on 02/Oct/21

a^{2} + {2 b}^{2} + {3 c}^{2} + {4 d}^{2} + {5 e}^{2}

= a^{2} + \frac{b^{2}}{\frac{1}{2}} + \frac{c^{2}}{\frac{1}{3}} + \frac{c^{2}}{\frac{1}{4}} + \frac{e^{2}}{\frac{1}{5}}

⩾ \frac{{(a + b + c + d + e)}^{2}}{1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}}

= \frac{1500}{137}

inequality hold only \frac{a}{1} = \frac{b}{\frac{1}{2}} = \frac{c}{\frac{1}{3}} = \frac{d}{\frac{1}{4}} = \frac{e}{\frac{1}{5}}

ie . a = \frac{300}{137} b = \frac{150}{137} c = \frac{100}{137} d = \frac{75}{137} e = \frac{60}{137}

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

t h a n k s s i r!

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

u s i n g m e t h o d a s i n Q 155495 :

S = a^{2} + 2 b^{2} + 3 c^{2} + 4 d^{2} + 5 e^{2}

= a^{2} + {(\frac{b}{\frac{1}{\sqrt{2}}})}^{2} + {(\frac{c}{\frac{1}{\sqrt{3}}})}^{2} + {(\frac{d}{\frac{1}{\sqrt{4}}})}^{2} + {(\frac{e}{\frac{1}{\sqrt{5}}})}^{2}

= {(a^{'})}^{2} + {(b^{'})}^{2} + {(c^{'})}^{2} + {(d^{'})}^{2} + {(e^{'})}^{2}

a + b + c + d + e = 5

a + \frac{1}{\sqrt{2}} \times (\frac{b}{\frac{1}{\sqrt{2}}}) + \frac{1}{\sqrt{3}} \times (\frac{c}{\frac{1}{\sqrt{3}}}) + \frac{1}{\sqrt{4}} \times (\frac{d}{\frac{1}{\sqrt{4}}}) + \frac{1}{\sqrt{5}} \times (\frac{e}{\frac{1}{\sqrt{5}}}) = 5

\Rightarrow a^{'} + \frac{b^{'}}{\sqrt{2}} + \frac{c^{'}}{\sqrt{3}} + \frac{d^{'}}{\sqrt{4}} + \frac{e^{'}}{\sqrt{5}} = 5

S_{m i n} = {(\frac{5}{\sqrt{1^{2} + {(\frac{1}{\sqrt{2}})}^{2} + {(\frac{1}{\sqrt{3}})}^{2} + {(\frac{1}{\sqrt{4}})}^{2} + {(\frac{1}{\sqrt{5}})}^{2}}})}^{2}

= \frac{25}{1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} = \frac{1500}{137}

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