10. Given $f(x)=x^{5}-10 x^{3}+a x^{2}+b x+c$. If the roots of the equation $f(x)=0$ are all real, and $m$ is the largest of these five real roots, then the maximum value of $m$ is $\qquad$
10. 4 .

Let the five real roots of $f(x)=0$ be
$$
x_{1} \leqslant x_{2} \leqslant x_{3} \leqslant x_{4} \leqslant m \text {. }
$$

By Vieta's formulas, we have
$$
\begin{array}{l}
m+x_{1}+x_{2}+x_{3}+x_{4}=0, \\
m \sum_{i=1}^{4} x_{i}+\sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}=-10 .
\end{array}
$$

Thus, $\sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}=-10+m^{2}$.
Then $\sum_{i=1}^{4} x_{i}^{2}=\left(\sum_{i=1}^{4} x_{i}\right)^{2}-2 \sum_{1 \leqslant i<j \leqslant 4} x_{i} x_{j}$.
$$
=m^{2}-2\left(-10+m^{2}\right)=20-m^{2} \text {. }
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{4} x_{i}\right)^{2} \leqslant 4 \sum_{i=1}^{4} x_{i}^{2} . \\
\text { Hence } m^{2} \leqslant 4\left(20-m^{2}\right) \\
\Rightarrow m^{2} \leqslant 16 \Rightarrow m \leqslant 4 . \\
\text { Also, for } f(x)=(x-4)(x+1)^{4} \\
=x^{5}-10 x^{3}-20 x^{2}-15 x-4,
\end{array}
$$

the equation $f(x)=0$ has all real roots, and the largest of the five real roots $m=4$.
Therefore, the maximum value of $m$ is 4.