For what of $n$ do we have $ a^n\le b^n+c^n$

Question

This is an inequality problem which I encountered while solving a metric space problem

let $a,b,c$ be non-negative real numbers satisfying $a\le b+c$.

Then for what positive real values of $n$ is the following also true

$$ a^n\le b^n+c^n$$

I honestly have no idea on how to proceed. I came up with this problem while solving a metric space question. Any hints and suggestions will be highly appreciated.

Answer 1

Note, the following sharps convex inequalities

For $p\ge 1$, $x\mapsto x^p$ is convex, then, $$ \left(\frac{a+b}{2}\right)^p \le \frac{1}{2}(a^p+b^p)\implies (a+b)^p\le 2^{p-1}(a^p+b^p)~~~~for~~~~p\ge 1$$ for $0< p<1$ $$(a+b)^p\ge a^p+b^p~~~~for~~~~p< 1$$

Hence $$a\le b+c \implies a^n\le 2^{n-1}(b^n+c^n)~~~~for~~~~n\ge 1$$

Answer 2

For $n>1$ it's obviously wrong.

We'll prove that it's true for all $0<n\leq1$.

Indeed, we need to prove that $$b^n+c^n\geq(b+c)^n,$$ which follows from Karamata (https://en.wikipedia.org/wiki/Karamata%27s_inequality):

Let $b\geq c$.

Thus, $(b+c,0)\succ(b,c)$ and since $f(x)=x^n$ is concave function, we obtain: $$b^n+c^n\geq(b+c)^n+0^n$$ and we are done!

We can use also the following reasoning.

$$\left(\frac{b}{b+c}\right)^n+\left(\frac{c}{b+c}\right)^n\geq\frac{b}{b+c}+\frac{c}{b+c}=1.$$

Answer 3

For $0<n<1:$

We have $0\leq a\leq b+c\implies a^n\leq (b+c)^n.$ So it suffices to prove $(b+c)^n\leq b^n+c^n.$

The case $\max (b,c)=0 $ is trivial, so for the non-trivial case, WLOG let $b>0.$ Dividing by $b^n$ and letting $c/b=d$ we have $$(b+c)^n\leq b^n+c^n\iff(1+d)^n\leq 1+d^n\iff$$ $$\iff \frac {1}{n}((1+d)^n-1)\leq \frac {1}{n}d^n\iff$$ $$\iff \int_1^{1+d}x^{n-1} dx\leq \int_0^dx^{n-1}dx \iff$$ $$\iff \int_0^d(x+1)^{n-1}dx\leq \int_0^dx^{n-1}dx\iff$$ $$\iff 0\leq\int_0^d (x^{n-1}-(x+1)^{n-1})dx.$$ Now for $0<n<1$ and $x>0$ the function $f(x)=x^{1-n}$ is decreasing. And $d\ge 0.$ So the last integral above is not negative.