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Question Number 186732 by mr W last updated on 09/Feb/23

	Answered by mr W last updated on 09/Feb/23

	${l e t}^{'} s g e n e r a l l y f i n d t h e m a x i m u m o f$ $S = a_{1} + a_{2} + . . . + a_{n} = \underset{k = 1}{\sum^{n}} a_{k}$ $w i t h$ ${(a_{1})}^{2} + {(2 a_{2})}^{2} + . . . + {({n a}_{n})}^{2} = \underset{k = 1}{\sum^{n}} {({k a}_{k})}^{2} = T$ $F = a_{1} + a_{2} + . . . + a_{n} + λ [{(a_{1})}^{2} + {(2 a_{2})}^{2} + . . . + {({n a}_{n})}^{2} - T]$ $\frac{\partial F}{\partial a_{k}} = 1 + 2 λ k^{2} a_{k} = 0$ $\Rightarrow a_{k} = - \frac{1}{2 λ k^{2}}$ $\underset{k = 1}{\sum^{n}} {({k a}_{k})}^{2} = T$ $\underset{k = 1}{\sum^{n}} {(- \frac{1}{2 λ k})}^{2} = T$ $\frac{1}{4 λ^{2}} \underset{k = 1}{\sum^{n}} \frac{1}{k^{2}} = T$ $\Rightarrow λ = \pm \frac{1}{2} \sqrt{\frac{\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}}{T}}$ $\Rightarrow a_{k} = \pm \frac{1}{k^{2}} \sqrt{\frac{T}{\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}}}$ $f o r a_{k} > 0, a_{k} = \frac{1}{k^{2}} \sqrt{\frac{T}{\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}}}$ $\Rightarrow S_{m a x} = \underset{k = 1}{\sum^{n}} a_{k} = \sqrt{\frac{T}{\underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}}} \underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}$ $= \sqrt{T \underset{k = 1}{\sum^{n}} \frac{1}{k^{2}}}$ $i n c u r r e n t c a s e : n = 50, T = 42925$ $\Rightarrow S_{m a x} = \sqrt{42925 \underset{k = 1}{\sum^{50}} \frac{1}{k^{2}}}$ $\approx 264.119$
	Answered by qaz last updated on 09/Feb/23

	$a_{1} + a_{2} + . . . + a_{50}$ $= \frac{1}{1} \cdot (1 \cdot a_{1}) + \frac{1}{2} (2 a_{2}) + \frac{1}{3} (3 a_{3}) + . . . + \frac{1}{50} (50 a_{50})$ $⩽ \sqrt{{(\frac{1}{1})}^{2} + {(\frac{1}{2})}^{2} + . . . + {(\frac{1}{50})}^{2}} \cdot \sqrt{{(1 \cdot a_{1})}^{2} + {(2 a_{2})}^{2} + . . . + {(50 a_{50})}^{2}}$ $= \sqrt{42925 \underset{k = 1}{\sum^{50}} \frac{1}{k^{2}}}$
	Commented by qaz last updated on 09/Feb/23

	$p r o v e :: a_{1} b_{1} + a_{2} b_{2} + . . . + a_{n} b_{n} ⩽ \sqrt{a_{1}^{2} + a_{2}^{2} + . . . + a_{n}^{2}} \cdot \sqrt{b_{1}^{2} + b_{2}^{2} + . . . + b_{n}^{2}} ? ? ?$ $\vec{a} = (a_{1}, a_{2}, . . ., a_{n}), \vec{b} = (b_{1}, b_{2}, . . ., b_{n})$ $\vec{a} \cdot \vec{b} =∣ \vec{a} ∣∣ \vec{b} ∣ \cos (\vec{a}, \vec{b})$ $\Rightarrow \vec{a} \cdot \vec{b} ⩽∣ \vec{a} ∣∣ \vec{b} ∣$ $\Rightarrow a_{1} b_{1} + a_{2} b_{2} + . . . + a_{n} b_{n} ⩽ \sqrt{a_{1}^{2} + a_{2}^{2} + . . . + a_{n}^{2}} \cdot \sqrt{b_{1}^{2} + b_{2}^{2} + . . . + b_{n}^{2}}$
	Commented by mr W last updated on 09/Feb/23

	$n i c e a p p r o a c h!$
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