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$$(\frac{n+1}{n})^n < (\frac{n}{n-1})^n$$

To prove the first case we'll replace n=2

LHS$\Rightarrow \frac{9}{4}$

RHS$\Rightarrow 4$

We can clearly see that LHS is less than RHS therefore first case is proven

Now we must assume that $n=k$ and $k\geq2$

$$(\frac{k+1}{k})^k < (\frac{k}{k-1})^k$$

Now we will have to prove for $n=k+1$

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

What i've tried to do from this point is working from the induction hypothesis

$(\frac{k+1}{k})^k < (\frac{k}{k-1})^k $ multiplying both sides by $(\frac{k+1}{k})$

$\Rightarrow (\frac{k+1}{k})^{k+1} < (\frac{k}{k-1})^k(\frac{k+1}{k})$

But that doesn't get me any closer. If anyone can help it'll be cool

Vicente Martinez
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    I suggest rewriting it as $$\left(1+\frac1n\right)^n<\left(1+\frac1{n-1}\right)^n$$ and comparing $1+\frac1n$ with $1+\frac1{n-1}$. – Brian M. Scott Apr 28 '20 at 02:22
  • Along similar lines, you could also cross multiply to get the equivalent inequality $(n^2 - 1)^n < (n^2)^n$. – user771918 Apr 28 '20 at 02:24

4 Answers4

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As suggested:

$$ ((n+2)n)^{n+1}=(n^2+2n)^{n+1}<(n^2+2n+1)^{n+1}=((n+1)^2)^{n+1} $$

which straightforwardly proves the inductive (and actually all) steps.

Brethlosze
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well we can first observe that the power on both sides are equal and is a positive number greater than or equal to 2.

now clearly (n+1)/n≤ 1≤ (n-1)/n now raising a power n≥2 the inequality sign remains unchanged,

hence {(n+1)/n}^n < {(n-1)/n}^n.

Ginger bread
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$(n^2-1)^{n}<(n^2)^n$ $\Rightarrow ((n-1)(n+1))^n<n^n \Rightarrow (\frac{n+1}{n})^n<(\frac{n}{n-1})^n$

We also get $(\frac{n-1}{n})^n<(\frac{n}{n+1})^n$.

Alapan Das
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We know that $n^2-1\leq n^2$ which implies $(n+1)(n-1)\leq n^2$ which further gives $\frac{n+1}{n}\leq \frac{n}{n-1}.$ Raising to the power $n$ on both the sides you get the result.

user159888
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