a_1 ,a_2 ,a_3 ,....,a_n >0 such that a_i ∈[0,i] ∀ i∈{1,2,3,4,...,n} prove that 2^n .a_1 (a_1 +a_2 )...(a_1 +a_2 +...+a_n )≥(n+1)(a_1 ^2 .a_2 ^2 ...a


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Question Number 194634 by York12 last updated on 12/Jul/23
$a_1 ,a_2 ,a_3 ,....,a_n >0 such that a_i ∈[0,i] ∀ i∈{1,2,3,4,...,n} prove that 2^n .a_1 (a_1 +a_2 )...(a_1 +a_2 +...+a_n )≥(n+1)(a_1 ^2 .a_2 ^2 ...a_n ^2 )$
$a_{1}, a_{2}, a_{3}, . . . ., a_{n} > 0 s u c h t h a t a_{i} \in [0, i]$ $\forall i \in {1, 2, 3, 4, . . ., n} p r o v e t h a t$ $2^{n} . a_{1} (a_{1} + a_{2}) . . . (a_{1} + a_{2} + . . . + a_{n}) ⩾ (n + 1) (a_{1}^{2} . a_{2}^{2} . . . a_{n}^{2})$
	Answered by York12 last updated on 12/Jul/23

	$a_{1} + a_{2} + a_{3} + . . . + a_{k} = 1 \times \frac{a_{1}}{1} + 2 \times \frac{a_{2}}{2} + 3 \times \frac{a_{3}}{3} + . . . k \times \frac{a_{k}}{k}$ $\frac{\underset{i = 1}{\sum^{k}} (\frac{i \times a_{i}}{i})}{\underset{i = 1}{\sum^{k}} (i)} ⩾ \underset{i = 1}{\prod^{k}} ({(\frac{a_{i}}{i})}^{\frac{i}{\underset{i = 1}{\sum^{k}} (i)}})$ $\Rightarrow \underset{i = 1}{\sum^{k}} (\frac{i \times a_{i}}{i}) ⩾ \underset{i = 1}{\sum^{k}} (i) \times \underset{i = 1}{\prod^{k}} ({(\frac{a_{i}}{i})}^{\frac{i}{\underset{i = 1}{\sum^{k}} (i)}})$ $\Rightarrow \underset{k = 1}{\prod^{n}} (\underset{i = 1}{\sum^{k}} (\frac{i \times a_{i}}{i})) ⩾ \underset{k = 1}{\prod^{n}} (\underset{i = 1}{\sum^{k}} (i) \times \underset{i = 1}{\prod^{k}} ({(\frac{a_{i}}{i})}^{\frac{i}{\underset{i = 1}{\sum^{k}} (i)}})$ $= \underset{k = 1}{\prod^{n}} (\underset{i = 1}{\sum^{k}} (i)) \times \underset{k = 1}{\prod^{n}} (\underset{i = 1}{\prod^{k}} ({(\frac{a_{i}}{i})}^{\frac{i}{\underset{i = 1}{\sum^{k}} (i)}})$ $= \frac{\underset{k = 1}{\prod^{n}} ((k) (k + 1))}{\underset{k = 1}{\prod^{n}} (2)} \times \underset{k = 1}{\prod^{n}} ({(\frac{a_{k}}{k})}^{(2 k \underset{m = k}{\sum^{n}} (\frac{1}{m (m + 1)}))})$ $\frac{1}{m (m + 1)} = \frac{1}{m} - \frac{1}{m + 1} \Rightarrow \underset{m = k}{\sum^{n}} (\frac{1}{m (m + 1)}) = \underset{m = k}{\sum^{n}} (\frac{1}{m}) - \underset{m = k}{\sum^{n}} (\frac{1}{m + 1}) = (\frac{1}{k} - \frac{1}{n + 1})$ $\frac{\underset{k = 1}{\prod^{n}} ((k) (k + 1))}{\underset{k = 1}{\prod^{n}} (2)} \times \underset{k = 1}{\prod^{n}} ({(\frac{a_{k}}{k})}^{(2 k \underset{m = k}{\sum^{n}} (\frac{1}{m (m + 1)}))})$ $= \frac{\underset{k = 1}{\prod^{n}} ((k) (k + 1))}{\underset{k = 1}{\prod^{n}} (2)} \times \underset{k = 1}{\prod^{n}} ({(\frac{a_{k}}{k})}^{(2 k (\frac{1}{k} - \frac{1}{n + 1})})$ $2 k (\frac{1}{k} - \frac{1}{n + 1}) = 2 (\frac{n + 1 - k}{n + 1})$ $(\frac{n + 1 - k}{n + 1}) ⩽ 1 \Rightarrow 2 (\frac{n + 1 - k}{n + 1}) ⩽ 2$ $(\frac{a_{k}}{k}) ⩽ 1 \Rightarrow {(\frac{a_{k}}{k})}^{2 (\frac{n + 1 - k}{n + 1})} ⩾ {(\frac{a_{k}}{k})}^{2}$ $\frac{\underset{k = 1}{\prod^{n}} ((k) (k + 1))}{\underset{k = 1}{\prod^{n}} (2)} \times \underset{k = 1}{\prod^{n}} ({(\frac{a_{k}}{k})}^{(2 k (\frac{1}{k} - \frac{1}{n + 1})}) ⩾ \frac{\underset{k = 1}{\prod^{n}} ((k) (k + 1))}{\underset{k = 1}{\prod^{n}} (2)} \times \underset{k = 1}{\prod^{n}} ({(\frac{a_{k}}{k})}^{(2)})$ $= \frac{n + 1}{2^{n}} (a_{1}^{2} a_{2}^{2} a_{3}^{2} . . . a_{n}^{2})$ $\Rightarrow \underset{k = 1}{\prod^{n}} (2 \underset{i = 1}{\sum^{k}} (\frac{i \times a_{i}}{i})) ⩾ (n + 1) (a_{1}^{2} a_{2}^{2} a_{3}^{2} . . . a_{n}^{2})$ $▽$
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