jmc

Squaring the equation $x+y+z=5,$ we get $$x^2+y^2+z^2+2(xy+xz+yz)=25.$$Then $x^2+y^2+z^2=25-2\cdot 8=9.$

By Cauchy-Schwarz, $$(1^2+1^2)(y^2+z^2)\ge (y+z{)}^2.$$Then $2(9-x^2)\ge (5-x{)}^2,$ which expands as $18-2x^2\ge 25-10x+x^2.$ This simplifies to $3x^2-10x+7\le 0,$ which factors as $(x-1)(3x-7)\le 0.$ Hence, $x\le \frac{7}{3}.$

Equality occurs when $y=z=\frac{4}{3},$ so the maximum value of $x$ is $\boxed{\frac{7}{3}}.$