Prove-the-following-inequalities-1-n-1-2-n-gt-n-for-n-N-n-gt-1-2-sinnx-n-sinx-for-n-N-

Question Number 118712 by 1549442205PVT last updated on 19/Oct/20

Prove the following inequalities :

1) {(\frac{n + 1}{2})}^{n} > n! for \forall n \in N^{*}, n > 1

2) ∣ sinnx ∣⩽ n ∣ sinx ∣ for \forall n \in N^{*}

Answered by Dwaipayan Shikari last updated on 19/Oct/20

\frac{1 + 2 + 3 + 4 + \dots n}{n} ⩾ {(1.2.3.4 . . n)}^{\frac{1}{n}}

\frac{n (n + 1)}{2 n} ⩾ {(n!)}^{\frac{1}{n}}

{(\frac{n + 1}{2})}^{n} ⩾ n!

t a k e n > 1

{(\frac{3}{2})}^{2} ⩾ 2! (W h i c h i s t r u e)

S o

{(\frac{n + 1}{2})}^{n} > n! (n > 1)

Commented by 1549442205PVT last updated on 20/Oct/20

Thank Sir .

Answered by TANMAY PANACEA last updated on 19/Oct/20

2) a p p l y i n g l o g i c

1 ⩾∣ s i n n x ∣⩾ 0

a n d 1 ⩾∣ s i n x ∣⩾ 0

b u t n ⩾ n ∣ s i n x ∣⩾ 0

s o n ∣ s i n x ∣>∣ s i n n x ∣

Commented by 1549442205PVT last updated on 19/Oct/20

I {don}^{'} t see logic here, {havn}^{'} t final

result n ∣ sinx ∣⩾∣ \sin (nx) ∣

Commented by TANMAY PANACEA last updated on 19/Oct/20

o k s i r i a m t r y i n g

Commented by 1549442205PVT last updated on 20/Oct/20

Thank Sir .

Answered by mindispower last updated on 19/Oct/20

1) \Leftrightarrow n l n (\frac{n + 1}{2}) ⩾ l n (\underset{k = 1}{\prod^{n}} k)

\underset{k = 1}{\prod^{n}} (k) ⩽ {(\frac{\underset{k = 1}{\sum^{n}} k}{n})}^{n}, A M - G M

\Rightarrow l n (\underset{k = 1}{\prod^{n}} k) ⩽ n l n (\frac{\underset{k = 1}{\sum^{n}} k}{n}) = n l n (\frac{n + 1}{2})

Commented by 1549442205PVT last updated on 20/Oct/20

Thank you sir

Answered by 1549442205PVT last updated on 20/Oct/20

We prove by the induction method

1 - For n = 2 we obtain the true inequality

2 < 9 / 4 . Suppose that k! < {(\frac{k + 1}{2})}^{k} . Then

by the induction hypothesis (k + 1)! =

k! (k + 1) < {(\frac{k + 1}{2})}^{k} (k + 1) . If now we prove

that {(\frac{k + 1}{2})}^{k} (k + 1) < {(\frac{k + 2}{2})}^{k + 1} (1)

the theorem is proved because then

(k + 1)! < {(\frac{k + 1}{2})}^{k} (k + 1) < {(\frac{k + 2}{2})}^{k + 1}

that is our inequality holds true for n = k + 1

Inequality (1) can clearly be rewritten

as \frac{{(k + 2)}^{k + 1}}{{(k + 1)}^{k + 1}} > \frac{2^{k + 1}}{2^{k}} or {(1 + \frac{1}{k + 1})}^{k + 1} > 2

But the binomial theorem yields

{(1 + \frac{1}{k + 1})}^{k + 1} = 1 + (k + 1) \frac{1}{k + 1} + \dots > 2

so the inequality (1) holds and thus the

oriinal inequality is proved

2) The inequality is obviously true for

n = 1 . Assuming that ∣ sinkx ∣⩽ k ∣ sinx ∣

, we prove that ∣ \sin (k + 1) x ∣⩽ (k + 1) ∣ sinx ∣

Indeed, using the inequality ∣ coskx ∣⩽ 1

we have ∣ \sin (k + 1) x ∣=∣ sinkx . cosx + sinxcoskx ∣

⩽∣ sinkx ∣ . ∣ cosx ∣ + ∣ sinx ∣∣ coskx ∣⩽

∣ sinkx ∣ + ∣ sinx ∣⩽ k ∣ sinx ∣ + ∣ sinx ∣

= (k + 1) ∣ sinx ∣ which shows the required

is true . The proof completed .