Prove the following inequalities: 1)(((n+1)/2))^n >n! for ∀n∈N^∗ ,n>1 2)∣sinnx∣≤n∣sinx∣ for ∀n∈N^∗


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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

	$\frac{1 + 2 + 3 + 4 + . . . n}{n} ⩾ {(1.2.3.4 . . n)}^{\frac{1}{n}}$ $\frac{n (n + 1)}{2 n} ⩾ {(n!)}^{\frac{1}{n}}$ ${(\frac{n + 1}{2})}^{n} ⩾ n!$ $t a k e n > 1$ ${(\frac{3}{2})}^{2} ⩾ 2! (W h i c h i s t r u e)$ $S o$ ${(\frac{n + 1}{2})}^{n} > n! (n > 1)$
	Commented by1549442205PVT 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 by1549442205PVT last updated on 19/Oct/20

	$I {don}^{'} t see logic here, {havn}^{'} t final$ $result n ∣ sinx ∣⩾∣ \sin (nx) ∣$
	Commented byTANMAY PANACEA last updated on 19/Oct/20

	$o k s i r i a m t r y i n g$
	Commented by1549442205PVT 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 by1549442205PVT last updated on 20/Oct/20

	$Thank you sir$
	Answered by 1549442205PVT last updated on 20/Oct/20

	$We prove by the induction method$ $1 - For n = 2 we obtain the true inequality$ $2 < 9 / 4 . Suppose that k! < {(\frac{k + 1}{2})}^{k} . Then$ $by the induction hypothesis (k + 1)! =$ $k! (k + 1) < {(\frac{k + 1}{2})}^{k} (k + 1) . If now we prove$ $that {(\frac{k + 1}{2})}^{k} (k + 1) < {(\frac{k + 2}{2})}^{k + 1} (1)$ $the theorem is proved because then$ $(k + 1)! < {(\frac{k + 1}{2})}^{k} (k + 1) < {(\frac{k + 2}{2})}^{k + 1}$ $that is our inequality holds true for n = k + 1$ $Inequality (1) can clearly be rewritten$ $as \frac{{(k + 2)}^{k + 1}}{{(k + 1)}^{k + 1}} > \frac{2^{k + 1}}{2^{k}} or {(1 + \frac{1}{k + 1})}^{k + 1} > 2$ $But the binomial theorem yields$ ${(1 + \frac{1}{k + 1})}^{k + 1} = 1 + (k + 1) \frac{1}{k + 1} + . . . > 2$ $so the inequality (1) holds and thus the$ $oriinal inequality is proved$ $2) The inequality is obviously true for$ $n = 1 . Assuming that ∣ sinkx ∣⩽ k ∣ sinx ∣$ $, we prove that ∣ \sin (k + 1) x ∣⩽ (k + 1) ∣ sinx ∣$ $Indeed, using the inequality ∣ coskx ∣⩽ 1$ $we have ∣ \sin (k + 1) x ∣=∣ sinkx . cosx + sinxcoskx ∣$ $⩽∣ sinkx ∣ . ∣ cosx ∣ + ∣ sinx ∣∣ coskx ∣⩽$ $∣ sinkx ∣ + ∣ sinx ∣⩽ k ∣ sinx ∣ + ∣ sinx ∣$ $= (k + 1) ∣ sinx ∣ which shows the required$ $is true . The proof completed .$
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