a-1-a-2-a-3-a-n-gt-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-

Question Number 194634 by York12 last updated on 12/Jul/23

a_{1}, a_{2}, a_{3}, \dots ., a_{n} > 0 s u c h t h a t a_{i} \in [0, i]

\forall i \in {1, 2, 3, 4, \dots, n} p r o v e t h a t

2^{n} . a_{1} (a_{1} + a_{2}) \dots (a_{1} + a_{2} + \dots + a_{n}) ⩾ (n + 1) (a_{1}^{2} . a_{2}^{2} \dots a_{n}^{2})

Answered by York12 last updated on 12/Jul/23

a_{1} + a_{2} + a_{3} + \dots + a_{k} = 1 \times \frac{a_{1}}{1} + 2 \times \frac{a_{2}}{2} + 3 \times \frac{a_{3}}{3} + \dots 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} \dots 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} \dots a_{n}^{2})

▽

Facebook Tweet Pin