Prove-that-a-1-a-2-a-n-n-a-1-2-a-2-2-a-n-2-n-with-equality-holding-iff-a-1-a-2-a-n-

Question Number 4719 by prakash jain last updated on 29/Feb/16

Prove that

\frac{a_{1} + a_{2} + \dots + a_{n}}{n} ⩽ \sqrt{\frac{a_{1}^{2} + a_{2}^{2} + \dots + a_{n}^{2}}{n}}

with equality holding iff a_{1} = a_{2} = \dots = a_{n} .

Commented by Yozzii last updated on 29/Feb/16

u = (a_{1}, a_{2}, a_{3}, \dots, a_{n}), m = \frac{1}{n} (1, 1, 1, \dots, 1)

\Rightarrow u . m = \frac{1}{n} \underset{r = 1}{\sum^{n}} a_{r}

∣ u ∣= \sqrt{\underset{r = 1}{\sum^{n}} a_{r}^{2}}, ∣ m ∣= \frac{1}{n} \sqrt{\underset{i = 1}{\sum^{n}} 1} = \frac{1}{\sqrt{n}}

∴ s i n c e c o s α = \frac{u . m}{∣ u ∣∣ m ∣} a n d ∣ c o s α ∣⩽ 1 (- π < α ⩽ π)

\Rightarrow∣ u . m ∣⩽∣ u ∣∣ m ∣

∴ ∣ \frac{1}{n} \underset{r = 1}{\sum^{n}} a_{r} ∣⩽ \sqrt{\frac{\underset{r = 1}{\sum^{n}} a_{r}^{2}}{n}}

T h i s i m p l i e s t h a t

- \sqrt{\frac{1}{n} (\underset{r = 1}{\sum^{n}} a_{r}^{2})} ⩽ \frac{1}{n} \underset{r = 1}{\sum^{n}} a_{r} ⩽ \sqrt{\frac{1}{n} (\underset{r = 1}{\sum^{n}} a_{r}^{2})}

F o r a l l a_{r} \in R w e t h e n d e d u c e

\frac{1}{n} \underset{r = 1}{\sum^{n}} a_{r} ⩽ \sqrt{\frac{1}{n} (\underset{r = 1}{\sum^{n}} a_{r}^{2})}

I f a_{1} = a_{2} = \dots = a_{n}

\Rightarrow \frac{1}{n} \times {n a}_{1} ⩽ \sqrt{\frac{1}{n} \times {n a}_{1}^{2}}

a_{1} ⩽ a_{1} \Rightarrow e q u a l i t y

I f e q u a i l t y o c c u r s

\Rightarrow \frac{1}{n^{2}} {(Σ a)}^{2} = \frac{1}{n} (Σ a^{2})

{(Σ a)}^{2} = n Σ a^{2}

F r o m t h i s w e o b t a i n c o s α = 1 \Rightarrow α = 0 .

\Rightarrow u = m \Rightarrow a_{1} = a_{2} = a_{3} = \dots = a_{n} = \frac{1}{n} .

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