7

Let $k$ be a positive integer and $x$ a positive real number. Prove that $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$.

This looks similar to Bernoulli's inequality. If we write $X=x^k$, the inequality is equivalent to $(1+X)^{\frac{k+1}{k}}\geq 1+X^{\frac{k+1}{k}}$, but this is not exactly in the form where we can apply Bernoulli.

Michael Rozenberg
  • 203,855
user336268
  • 2,439
  • What assumptions of the real numbers are we allowing? Because certain assumptions make this trivially easy. – Rushabh Mehta Aug 08 '18 at 21:29
  • $$\begin{align}(1+x^k)^{k+1} & = (1+x^k)^k(1+x^k) \ & = (1+x^k)^k + x^k(1+x^k)^k \ & = (1+x^k)^k + (x+x^{k+1})^k\end{align}$$ Not sure if this helps, but it may. – SlipEternal Aug 08 '18 at 21:29
  • It all depends on the assumptions of the properties of the real numbers that he's allowed to assume. This could either be really easy or really hard. – Rushabh Mehta Aug 08 '18 at 21:30
  • It looks like a modified triangle inequality... – SlipEternal Aug 08 '18 at 21:31

4 Answers4

7

Let $f(X) = (1+X)^{r} - 1 - X^{r}$ for $r> 1$. Then $f(0)=0$ and $$ f'(X) = r\left((1+X)^{r-1}-X^{r-1}\right)\geq 0 $$ since $X\mapsto X^{r-1}$ is an increasing function. Hence $f$ is also increasing and we get $f(X)\geq f(0)=0$ for $X\geq 0$. Now set $r = (k+1)/k$.

Seewoo Lee
  • 15,670
3

Let $f(x)=x^{\alpha}$, where $\alpha>1$.

Thus, $f$ is a convex function and since for positives $a$ and $b$ such that $a\geq b$ we have $$(a+b,0)\succ (a,b),$$ by Karamata we obtain: $$f(a+b)+f(0)\geq f(a)+f(b)$$ or $$(a+b)^{\alpha}\geq a^{\alpha}+b^{\alpha}$$ and since the last inequality is symmetric, it's true for all positives $a$ and $b$.

Id est, for $a=1$, $b=x^{k}$ and $\alpha=\frac{k+1}{k}$ we got your inequality.

Michael Rozenberg
  • 203,855
1

Case 1. $x\le 1$.

Then $x^k\ge x^{k+1}$ and therefore $$(1+x^k)^{k+1} = (1+x^k)(1+x^k)^k \ge (1+x^k)^k \ge (1+x^{k+1})^k.$$

Case 2. $x>1$.

We divide the inequality by $x^{k(k+1)}$ and get the following equivalent form $$\left(1+\left(\frac 1x \right)^k\right)^{k+1} \ge \left(1+\left(\frac 1x \right)^{k+1}\right)^k.$$ This is just Case 1. for the number $\dfrac 1x$.

timon92
  • 12,291
0

For $r\geq 1$, we have by Minkowski's Inequality that $$X+1=\big(X^r+0^r\big)^{\frac{1}{r}}+\big(0^r+1^r\big)^{\frac1r}\geq \left((X+0)^r+(0+1)^r\right)^{\frac{1}{r}}=\left(1+X^r\right)^{\frac{1}{r}}\text{ for all }X\geq0\,.$$ This shows that $$(1+X)^r\geq 1+X^r\,.$$ Note that the inequality becomes an equality iff $r=1$ or $X=0$. Now, for $k,l\in\mathbb{R}_{>0}$ with $k\leq l$, we then see that, with $X:=x^k$ and $r:=\frac{l}{k}$, we have $$\left(1+x^k\right)^{\frac{l}{k}}\geq 1+x^l\,,\text{ or }\left(1+x^k\right)^l\geq\left(1+x^l\right)^k\,,$$ for all $x>0$. The equality case is when $x=0$ and when $k=l$. This problem is a particular case where $l=k+1$. Furthermore, we conclude that the function $f:\mathbb{R}_{\geq0}\times\mathbb{R}_{>0}\to\mathbb{R}$ defined by $$f(x,k):=\left(1+x^k\right)^{\frac1k}\text{ for all }x\geq0\text{ and }k>0$$ is a nondecreasing function in $k$, and it is strictly increasing with respect to $k$ for $x>0$.

Batominovski
  • 50,341