Is it true that $(a+b)^x>a^x+b^x$ for three numbers $a,b,x>1$

Question

This holds for if $x$ is natural nunbers greater than $1$ so i think it would work for every real nunber $x$ greater than $1$.

Answer 1

Yes you are right. It's true for any reals $x>1$ and non-negatives $a$ and $b$.

Indeed, let $f(t)=t^x$, where $t\geq0$.

Thus, $f$ is a convex function.

We can assume that $a\geq b$.

Thus, since $$(a+b,0)\succ(a,b),$$ by Karamata we obtain: $$f(a+b)+f(0)\geq f(a)+f(b)$$ or $$(a+b)^x\geq a^x+b^x.$$

Answer 2

I am turning my previous comment into an answer.

Dividing both sides by $(a+b)^x$, the original inequality is equivalent to $$p^x+q^x<1,\quad \forall p>0, q>0 \ni p+q=1, x>1.$$

We now demonstrate the last statement.

$u^x$ decreases with respect to $x$ for $u\in(0,1)$. So $$p^x+q^x<p^1+q^1=1, \,\forall x>1;$$ and $$p^x+q^x<p^1+q^1=1,\,\forall x<1.$$

Answer 3

Here is a simple proof using the binomial theorem:

Let $n = \left\lfloor x \right\rfloor$, then we have that \begin{equation} (a+b)^x = (a+b)^{x-n }\sum_{k =0}^n \binom{n}{k } a^{n-k}b^{k}, \end{equation} and therefore it is enough to prove that \begin{equation} \sum_{k =0}^n \binom{n}{k } a^{n-k}b^{k} > \frac{a^x}{(a+b)^{x -n}} + \frac{b^x}{(a+b)^{x-n}}, \end{equation} If $x- n = 0$ then we are done. The proof is as trivial as noticing that $a^x$ and $b^x$ are two terms on the left. Therefore assume $x- n \neq 0$.

Towards that end notice that since $a,b>0$ we have that \begin{equation} \frac{a^{x-n}}{(a+b)^{x-n}} < 1 \end{equation} and likewise \begin{equation} \frac{b^{x -n}}{(a+b)^{x-n}} < 1 \end{equation} since $y^\alpha$ is a strictly increasing function of $y$. Therefore we have that \begin{equation} \frac{a^{x}}{(a+b)^{x-n}} + \frac{b^{x}}{(a+b)^{x-n}} < a^n + b^n \end{equation} which implies \begin{equation} \sum_{k =0}^n \binom{n}{k } a^{n-k}b^{k} \geq a^n +b^n >\frac{a^{x}}{(a+b)^{x-n}} + \frac{b^{x}}{(a+b)^{x-n}} , \end{equation} as was needed.