3. Given real numbers $a, b, c$ satisfy
$$
f(x)=a \cos x+b \cos 2 x+c \cos 3 x \geqslant-1
$$

for any real number $x$. Then the maximum value of $a+b+c$ is . $\qquad$
3.3.

Notice that,
$$
\begin{array}{l}
f(0)+f\left(\frac{\pi}{2}\right)+f(\pi)+f\left(\frac{3 \pi}{2}\right)=0 \\
\Rightarrow f(0)=-\left(f\left(\frac{\pi}{2}\right)+f(\pi)+f\left(\frac{3 \pi}{2}\right)\right) \leqslant 3 .
\end{array}
$$

When $a=\frac{3}{2}, b=1, c=\frac{1}{2}$, we have
$$
\begin{array}{l}
f(x)=\frac{3}{2} \cos x+\cos 2 x+\frac{1}{2} \cos 3 x \\
=2 \cos ^{3} x+2 \cos ^{2} x-1 \\
=2 \cos ^{2} x \cdot(\cos x+1)-1 \\
\geqslant-1 .
\end{array}
$$

Therefore, the maximum value of $a+b+c$ is 3.