Showing that for $x \ge 7$, $x\# \ge x^2+x$ Let $x\#$ be the primorial of $x$.
I am trying to show that if $x \ge 7$:

$$x\# \ge x^2+x$$

Is there a straight forward argument?
Here's what I came up with:
(1) From Bertrand's Postulate, for any $x$, there exists a prime $p$ such that $x < p < 2x$ 
(2) Base Case: $7\# = 210 \ge (14)^2 + (14) = 210$
(3) Assume for a prime $p \ge 7$, $p\# \ge (2p)^2 + (2p)$
(4) Let $p+c$ be the lowest prime greater than $p$ so that from step 1 above, $2 \le c < p$ 
(5) $(2p+2c)^2 + 2(p+c) = (4p^2+2p+2c) + 8pc + 4c^2 < (4p^2 + 4p) +  (8p^2 + 8p) + (4p^2 + 4p) < 2p\# + 4p\#+ 2p\# = (2+4+2)p\# < (p+c)p\#$

The idea is pretty good, but I feel the solution was difficult to read. I will just rephrase your ideas more clearly. 
1) Claim: It is enough to prove that $p\#\geq (2p)^2+2p$ for $p$ prime.
Indeed, let $x$ any natural number, then we can find  $p_{n}\leq x<p_{n+1}$. By Bertrand postulate, we have $ p_{n+1}<2p_n$. Hence,
$$x\#=p_{n}\#\geq (2p_n)^2+(2p_n)\geq x^2+x.$$
2) We are going to prove the claim by induction:
Case p=7 is easy to check.
Assume hypothesis for $p_n$, let's prove it for $p_{n+1}$: 
Recall $p_{n+1}\leq 2p_n$. Then, we have$$ (2p_{n+1})^2+(2p_{n+1})\leq (4p_n)^2+4p_n\leq 4((2p_n)^2+(2p_n))\leq 4p_n\#\leq p_{n+1}\#$$
Which is what we wanted to prove.
