Step 1: Recall the rule for decreasing.
A function decreases everywhere when its derivative is never positive, that is $f'(x)\le0$ for every real $x$.
Step 2: Differentiate the function.
Given $f(x)=\sqrt{3}\sin x-\cos x-2ax+b$. Differentiate term by term: \[ f'(x)=\sqrt{3}\cos x+\sin x-2a. \]
Step 3: Write the decreasing condition.
We need $\sqrt{3}\cos x+\sin x-2a\le0$ for all $x$, which means \[ \sqrt{3}\cos x+\sin x\le2a\quad\text{for every }x. \]
Step 4: Find the biggest value of the left side.
Any expression $p\cos x+q\sin x$ has greatest value $\sqrt{p^2+q^2}$. Here $p=\sqrt{3}$ and $q=1$, so the maximum is \[ \sqrt{3+1}=2. \]
Step 5: Make the inequality always hold.
For the inequality to be true even at the highest point, $2a$ must be at least this maximum value: \[ 2a\ge2. \]
Step 6: Solve for $a$.
Dividing both sides by $2$ gives $a\ge1$. So the function decreases everywhere exactly when $a\ge1$.
\[ \boxed{a\ge1} \]