which one is larger $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$? Which is larger? $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$?
Algebraic justilation does not help.
 A: $\sqrt[n]{x+\delta}-\sqrt[n]{x}\ \boxed{\phantom{A} }\sqrt[n]{x}-\sqrt[n]{x-\delta}$
$\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}\ \boxed{\phantom{A} }2\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{\frac{\left(\sqrt[n]{x+\delta}\right)^n+\left(\sqrt[n]{x-\delta}\right)^n}{2}}$
But $\frac{a+b}{2}\leq\sqrt[n]{\frac{a^n+b^n}{2}}$.
$\tiny{\text{some conditions apply}}$
A: The $n$-th root function is concave increasing on $(0,\infty)$. So its rate of change is decreasing. Hence the second quantity is larger.
Edited to add: Alfe has a nice picture that explains this clearly.
A: Let's generalize it even further:
For $\alpha>0$ and $x>\delta>0$, we want to decide which is bigger $\left(x+\delta\right)^\alpha-x^\alpha$ or $x^\alpha-\left(x-\delta\right)^\alpha$.
So, let $f(\alpha)=\left(\left(x+\delta\right)^\alpha-x^\alpha\right)-\left(x^\alpha-\left(x-\delta\right)^\alpha\right)$. We want to decide $\text{sign}(f(\alpha))$.
First:
$\begin{align} f(\alpha)
&=\left(x+\delta\right)^\alpha+\left(x-\delta\right)^\alpha-2x^\alpha-x^\alpha
\\
&=x^\alpha\left(1+\frac{\delta}{x}\right)^\alpha+x^\alpha\left(1-\frac{\delta}{x}\right)^\alpha-2x^\alpha
\\
&=x^\alpha\left(\left(1+\frac{\delta}{x}\right)^\alpha+\left(1-\frac{\delta}{x}\right)^\alpha-2\right)
\end{align}$
Note that $x^\alpha>0$, so we can divide by it, and $0<\rho:=\frac{\delta}{x}<1$. Therefore we only need to decide the sign of:
$g(\alpha)=\left(1+\rho\right)^\alpha+\left(1-\rho\right)^\alpha-2$
We notice that $g(1)=1+\rho+1-\rho-2=0$, and also that $g'(\alpha)=\alpha\left(1+\rho\right)^{\alpha-1}+\alpha\left(1-\rho\right)^{\alpha-1}>0$.
We conclude that $g$ is strictly increasing, thus $\text{sign}(f(\alpha))=\text{sign}(\alpha-1)$, which means:


*

*$\alpha<1 \Rightarrow \left(x+\delta\right)^\alpha-x^\alpha<x^\alpha-\left(x-\delta\right)^\alpha$

*$\alpha=1 \Rightarrow \left(x+\delta\right)^\alpha-x^\alpha=x^\alpha-\left(x-\delta\right)^\alpha$

*$\alpha>1 \Rightarrow \left(x+\delta\right)^\alpha-x^\alpha>x^\alpha-\left(x-\delta\right)^\alpha$


For your question, if $n>1$ then $\alpha=\frac{1}{n}<1$, and we have $\left(x+\delta\right)^\alpha-x^\alpha<x^\alpha-\left(x-\delta\right)^\alpha$.
A: Hint: Try using the Mean-Value-Theorem for derivatives.
But there is algebraic justification too, note that $$ a-b = \left(\sqrt[n]{a}-\sqrt[n]{b}\right) \times \left\{(\sqrt[n]{a})^{n-1}+(\sqrt[n]{a})^{n-2}\sqrt[n]{b}+\ldots+\sqrt[n]{a}(\sqrt[n]{b})^{n-2}+(\sqrt[n]{b})^{n-1}\right\} $$
A: How about using the fact that $(a-b)(\sum_{i=0}^{n-1}a^{n-1-i}b^i) = a^n - b^n$
A: We all know that when $n>1$ the curve $x\mapsto y:=x^n$ $\>(x\geq0)$ gets steeper and steeper as $x$ increases. It follows that the same curve when viewed as a curve over the $y$-axis, which is the same thing as viewing the graph of $y\mapsto x:={\root n\of{\mathstrut y}}$,  gets ever flatter as $y$ increases. This implies that for given  $\delta>0$ and $y\geq\delta$ one has
$${\root n\of {y+\delta}}-{\root n\of {\mathstrut y}}\leq {\root n\of {\mathstrut y}}-{\root n\of {y-\delta}}\ .$$
A: Actually there is an Algebraic Justification,
$\sqrt[n]{x+\delta}-\sqrt[n]{x}=\frac{\delta}{\left((\sqrt[n]{x+\delta})^{n-1}+(\sqrt[n]{x+\delta})^{n-2}(\sqrt[n]{x})+\cdots+(\sqrt[n]{x})^{n}\right)}$,
$\sqrt[n]{x}-\sqrt[n]{x-\delta}=\frac{\delta}{\left((\sqrt[n]{x})^{n-1}+(\sqrt[n]{x})^{n-2}(\sqrt[n]{x-\delta})+\cdots+(\sqrt[n]{x-\delta})^{n}\right)}$.
Comparing the denominators we see that $\sqrt[n]{x+\delta}-\sqrt[n]{x}\le\sqrt[n]{x}-\sqrt[n]{x-\delta}$
