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If m and n are positive integers then show that:$$\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$$I am new in this Course.So i can't able to think how i start a inequalities question by looking it's pattern.Can anyone help me to explain that Inequalities and Thanks in advance.

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  • $\begingroup$ Note that $n^m=n^m\cdot 1^1$. And the inequality should not be strict. The equality case is $(m,n)=(1,1)$. $\endgroup$ – Batominovski Aug 28 '18 at 17:01
  • $\begingroup$ Can you explain me what the equality case is (m,n)=(1,1) mean by.Sorry if the question is irrelevant.I couldn't understand it.please explain it further more details @Batominovski $\endgroup$ – emonHR Aug 28 '18 at 17:07
  • $\begingroup$ @mdemon when $m=1,n=1$, the LHS is $(2/2)^2$ and the RHS is $1^1$, which are both $1$ $\endgroup$ – John Glenn Aug 28 '18 at 17:11
  • $\begingroup$ @mdemon Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – user Sep 17 '18 at 20:20
  • $\begingroup$ Oops! i forgot @gimusi Thanks for mention it $\endgroup$ – emonHR Sep 20 '18 at 6:21
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As an alternative by induction we have

  • Base case

$$n=1 \implies \left(\frac{m+1}{m+1}\right)^{m+1} \ge 1^m$$

  • Induction step, assume $\left(\frac{mn+1}{m+1}\right)^{m+1} \ge n^m$ true we need to show that $\left(\frac{m(n+1)+1}{m+1}\right)^{m+1} \ge (n+1)^m$ then

$$\left(\frac{m(n+1)+1}{m+1}\right)^{m+1}=\left(\frac{mn+1}{m+1}+\frac{m}{m+1}\right)^{m+1}\ge \left(n^m+\frac{m}{m+1}\right)^{m+1}\ge n^{m(m+1)}\ge n^m$$

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Apply A.M. G.M. $$\frac{\underbrace{n+n+\cdots+n}_{m\ \text{times}}+1}{m+1}\ge (n^m\cdot 1)^{\frac{1}{m+1}}$$ $$\frac{mn+1}{m+1}\ge (n^m\cdot 1)^{\frac{1}{m+1}}$$ $$\left(\frac{mn+1}{m+1}\right)^{m+1}\ge n^m\cdot 1$$

$$\left(\frac{mn+1}{m+1}\right)^{m+1}\ge n^m$$

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  • $\begingroup$ Thanks a lot @James I really looking for this explanation.Can you suggest me a good book to start inequalities problem to build up my concept please. $\endgroup$ – emonHR Aug 28 '18 at 17:19
  • $\begingroup$ I don't know but you can ask it as new question on this site itself $\endgroup$ – Deepesh Meena Aug 28 '18 at 17:29
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Apply AM-GM inequality for $m$ numbers $n$ and $1$ number $1$ we have the result follows by raising both sides of the inequality to the $m+1$ power.

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So, I'm not fantastic at writing proofs, but I think you can use a bit of intuition to determine whether this is true.

$$\Biggl(\frac{mn + 1}{m + 1}\Biggl)^{m + 1} > n^m$$

The numerator $(mn + 1)^{m + 1}$ can be expressed as a polynomial of the form $ax^n + bx^{n - 1} + ... + c$. The highest order term in polynomials of that form is the leading term, $ax^n$. Therefore, in $(mn + 1)^{m + 1}$, the term $(mn)^{m + 1}$. Because it grows faster than all other terms, you can regard it to be an approximation of the result of the numerator. The same reasoning can be applied to the denominator. Therefore, we get the following expression:

$$\Biggl(\frac{(mn)^{m + 1}}{m^{m + 1}}\Biggl)$$

Which simplifies to

$$n^{m + 1}$$

Clearly, $$n^{m + 1} > n^m$$

Once again, this proof (if you can call it that?) is not really rigorous, but I think it's an intuitive way to prove it.

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