 Prime-counting upper bound from a primorial estimate

If $P(X)\le A^X$ for every $X\ge 1$, where $A>1$, thenfor $X\ge 2$. Every prime in $(\sqrt{X},X]$ contributes at least $\sqrt{X}$ to the primorial, soTaking logarithms, using $\pi (\sqrt{X})\le \sqrt{X}$, and observing that $\log X\le \sqrt{X}$ for $X\ge 2$ proves the claim.