Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.

It is trivial that for any $a>0$ and $b>0$ (just raise both side to the $n^{th}$ power)
$$a^\frac1n +b^\frac1n >(a+b)^\frac1n$$
therefore $\forall x>0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}> \sqrt[4]{2x+1}$$
and for $x=0$
$$\sqrt[4]{x}+\sqrt[4]{x+1}= \sqrt[4]{2x+1}=1$$
