How can I prove that $(a+b)^{1/p} > a^{1/p} +b^{1/p}$ for every $0< p< 1.$ and for every $a,b > 0.$

Question

How can I prove that $(a+b)^{1/p} > a^{1/p} +b^{1/p}$ for every $0< p< 1.$ and for every $a,b > 0.$

I can see this proof on this site Does $|x|^p$ with $0<p<1$ satisfy the triangle inequality on $\mathbb{R}$? but I do not understand how this will help me in my proof? could anyone give me a simple proof for my question.(because I do not quite understand the proof given in the above link.)

Answer 1

So $1/p>1$ and if we ler $r=1/p>1$ you are to prove $(a+b)^{r}>a^{r}+b^{r}$ for $a,b>0$. If suffices to prove $(1+x)^{r}>1+x^{r}$ for $x>0$, for you may substitute $x=b/a$.

Let $\varphi(x)=(1+x)^{r}-(1+x^{r})$ for $x>0$, then $\varphi(0)=0$ and $\varphi'(x)=r(1+x)^{r-1}-rx^{r-1}=r((1+x)^{r-1}-x^{r-1})>0$, as $1+x>x$, $(1+x)^{r-1}>x^{r-1}$.

So $\varphi$ is strictly increasing, this means that $\varphi(x)>\varphi(0)$, so $(1+x)^{r}-(1+x^{r})>0$ and hence $(1+x)^{r}>1+x^{r}$.

Answer 2

Consider two functions: $f(x)=(a+x)^{\frac{1}{p}}-a^{\frac{1}{p}}$ and $g(x)=x^{\frac{1}{p}}$. Notice, that $f(0)=g(0)=0$ and for every $x \geq 0$ $f'(x) > g'(x)$, because:

$$f'(x)=\frac{1}{p}(a+x)^{\frac{1}{p}-1}=\frac{1}{p}(a+x)^{\frac{1-p}{p}} > \frac{1}{p}(x)^{\frac{1-p}{p}} = g'(x)$$

This holds, because $1-p>0$ and $a>0$.

Concluding, $f(x)>g(x)$ for every $x>0$, which proves the inequality after substitution $x=b$.