# Inequality regarding sum and root

I need your expertise in order to solve the following problem:

Let $x \in \mathbb{R}$ and let $k \in (0,\infty)$ . It is known that $$\sqrt{1} + \sqrt{e^x} \leq \sqrt{2} \cdot \sqrt{1 + e^x}.$$ How can this be generalized to any root, i.e. $$\sqrt[k]{1} + \sqrt[k]{e^x} \leq \sqrt[k]{2} \cdot \sqrt[k]{1 + e^x} \ ?$$

Is it possible? If not what is the closest form of inequality that resembles the above inequality?

No. Set $x=0$, then $2\sqrt[k]2\leq(\sqrt[k]2)^2$ so $k\leq1$. What is true is
$$\sqrt[k]1+\sqrt[k]{e^x}\leq\frac2{\sqrt[k]2}\sqrt[k]{1+e^x}$$
• does the above inequality hold for $k \in (0,1)$ ? – user3492773 Sep 19 '16 at 7:06
for $k=10$ doesn't this inequality holds for example, set here $x=0$
You could differentiate the function $$f_k(x)=\dfrac{1+\sqrt[k]{e^x}}{\sqrt[k]{1+e^x}}$$ where $k\in\mathbb{N}$ is given. You obtain that $f_k$ always takes its maximum value when $x=0$ which is $$f_k(0)=\dfrac{2}{\sqrt[k]{2}}$$ hence $$1+\sqrt[k]{e^x}\leq\dfrac{2}{\sqrt[k]{2}}\sqrt[k]{1+e^x}$$