5. Solve the equation $\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}=\sqrt{x+1}$.
5. According to the Cauchy-Schwarz inequality
$$
\left(\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}\right)^{2} \leqslant\left[(\sqrt{1-\sqrt{x}})^{2}+(\sqrt[4]{x})^{2}\right] \cdot\left[1^{2}+(\sqrt{x})^{2}\right] \text {, }
$$

Since $\sqrt{1-\sqrt{x}}+\sqrt[4]{x^{3}}=\sqrt{x+1}$,
then $\quad\left(\sqrt{1-\sqrt{x}}+\sqrt[1]{x^{3}}\right)^{2}=\left[(\sqrt{1-\sqrt{x}})^{2}+(\sqrt[1]{x})^{2}\right] \cdot\left[1+(\sqrt{x})^{2}\right]=x+1$. According to the condition for equality in the Cauchy-Schwarz inequality, we have and only have
$$
\left\{\begin{array}{l}
\sqrt{1-\sqrt{x}}=k \cdot 1, \\
\sqrt[4]{x}=k \sqrt{x} \text { ( } k \text { is a constant). }
\end{array}\right.
$$

Solving this, we get the root of the original equation as $x=0$.