Prove-that-p-n-a-1-a-2-a-n-n-n-a-1-a-2-a-n-n-N-

Question Number 45353 by pieroo last updated on 12/Oct/18

Prove that p (n) = \frac{a_{1} + a_{2} + \dots + a_{n}}{n} ⩾^{n} \sqrt{a_{1} a_{2} \dots a_{n}}

\forall n \in N

Commented by pieroo last updated on 12/Oct/18

please help

Commented by Kunal12588 last updated on 12/Oct/18

t r y i n g P M I (I n d u c t i o n)

Commented by pieroo last updated on 14/Oct/18

I still need help urgently please

Answered by Kunal12588 last updated on 12/Oct/18

p (n) : \frac{a_{1} + a_{2} + a_{3} + \dots + a_{n}}{n} ⩾ \sqrt[n]{a_{1} a_{2} a_{3} \dots a_{n}}

p (1) : \frac{a_{1}}{1} = a_{1} \sqrt[1]{a_{1}} = a_{1}

∴ p (1) s a t i s f i e s p (n)

p (2) : \frac{a_{1} + a_{2}}{2} \sqrt[2]{a_{1} a_{2}} = \sqrt{a_{1}} \sqrt{a_{2}}

\frac{1}{2} {(\sqrt{a_{1}} - \sqrt{a_{2}})}^{2} ⩾ 0

\Rightarrow \frac{a_{1} + a_{2} - 2 \sqrt{a_{1}} \sqrt{a_{2}}}{2} ⩾ 0

\Rightarrow \frac{a_{1} + a_{2}}{2} ⩾ \sqrt{a_{1}} \sqrt{a_{2}}

∴ p (2) s a t i s f i e s p (n)

l e t u s a s s u m e p (n) i s t r u e f o r n = k

\Rightarrow p (k) : \frac{a_{1} + a_{2} + a_{3} + \dots + a_{k}}{k} ⩾ \sqrt[k]{a_{1} a_{2} a_{3} \dots a_{k}}

n o w w e h a v e t o s h o w

p (k + 1) : \frac{a_{1} + a_{2} + a_{3} + \dots + a_{k + 1}}{k + 1} ⩾ \sqrt[k + 1]{a_{1} a_{2} a_{3} \dots a_{k + 1}}

p l e a s e h e l p

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

i a m h e r e b y p o s t i n g p r o o f g i v e n i n H i g h e r A l g e b r s

B e r n a r d a n d c h i l d \dots

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18