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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.)

Intuition
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  • Adapt the proof of the accepted answer, i.e., construct functions, take the derivative to get the monotonicity of the function, hence an inequality of the function, and go back to your inequality. – xbh Dec 03 '19 at 16:32

2 Answers2

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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 $x=b$.

Andronicus
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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 $(1+x)^{r}>1+x^{r}$.

user284331
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