1.4.11 $\star \star$ Given the equation in terms of $x$
$$
\sqrt{x}+\sqrt[3]{x+7}=\sqrt[4]{x+80}+c
$$

has real solutions. Find the range of real values for $c$.
The problem is equivalent to: Find the range of the function $f(x)=\sqrt{x}+\sqrt[3]{x+7}-\sqrt[4]{x+80}$. The domain of $f(x)$ is $[0,+\infty)$, and on this interval, the function $y=\sqrt[3]{x+7}$ is an increasing function. Next, we examine the function $g(x)=\sqrt{x}-\sqrt[4]{x+80}(x \geqslant 0)$.
Notice that, when $x>0$, we have $g^{\prime}(x)=\frac{1}{2 \sqrt{x}}-\frac{1}{4 \cdot \sqrt[4]{(x+80)^{3}}}$.
And when $x>0$, we have $2 \sqrt{x}0$ implies $g^{\prime}(x)>0$. This indicates that: when $x \in[0,+\infty)$, $g(x)$ is also an increasing function. Therefore, $f(x)$ is increasing on $[0,+\infty)$. Thus, the equation has real solutions if and only if $c \in[f(0),+\infty)$. Therefore, the range of $c$ is:
$$
c \geqslant \sqrt[3]{7}-\sqrt[4]{80}=\sqrt[3]{7}-2 \sqrt[4]{5} .
$$