In triangle ABC $\cos A+\cos B+\cos C\le P$ then $P=$

For a function which is concave downwards

$\boxed{\frac{f\left(x_1\right)+f\left(x_2\right)+f\left(x_3\right)}{3}\le f\left(\frac{x_1+x_2+x_3}{3}\right)}$

$$\begin{gathered} \text{ Let }P(A,\cos A),Q(B,\cos B)\text{ and }R(C,\cos C) be 3 points on the curve \\ y=\cos x \end{gathered}$$

$\begin{array}{r}\frac{\cos A+\cos B+\cos C}{3}\le \cos \left(\frac{A+B+C}{3}\right)=\frac{1}{2}\\ \left\{\because A+B+C={180}^{\circ }\right\}\\ \therefore \cos A+\cos B+\cos C\le \frac{3}{2}\end{array}$