How to show that: $\left(\frac{mn+1}{m+1}\right)^{m+1} > n^m$

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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

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

Answer 4

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.