Answer
$(-\infty,\infty)$ or $\{x|x=all\ real\ numbers\}$.
Work Step by Step
Step 1. $6(x^2+1)\ge5x \longrightarrow 6x^2-5x+6\ge0 \longrightarrow b^2-4ac=(-5)^2-4(6)(6)=-119\lt0$.
Step 2. Thus the quadratic opens up and has no x-intercept, meaning the curve is above the x-axis and the solution interval is $(-\infty,\infty)$ or $\{x|x=all\ real\ numbers\}$.
