Show that if $n$ is is a positive integer such that $n\ne 2$ and $n\ne 6$ then $\phi(n) \ge \sqrt n$ $\phi(n)$ being Euler's totient function.
Regarding effort put into the problem:
In the case that $n$ is a prime $p$, then it is given that $\phi(p) = p-1$. It is also given that $n\ne 2$, so the fact that $p-1 \ge \sqrt p$, or $(p-1)^2 \ge p$ is easy enough to prove in it of itself.
Considering the case were $n$ is some non-prime number, then $n$ may be written as its power prime factorization $n=p^{a_1}_1p^{a_2}_2 \ldots p^{a_k}_k$. It is given that $\phi$ is multiplicative, so applying this to the original problem we have $\phi(n)=\phi(p^{a_1}_1) \phi(p^{a_2}_2) \ldots \phi(p^{a_k}_k)$.
It is given that $\phi(p^{a_j}_j)=p^j_j-p^{a_j-1}_j=p^{a_j}_j(1-\frac{1}{p_j})$, so the above can be rewritten as
$\phi(n)=p^{a_1}_1p^{a_2}_2 \ldots p^{a_k}_k(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \ldots (1-\frac{1}{p_k})=n(1-\frac{1}{p_1})(1-\frac{1}{p_2}) \ldots (1-\frac{1}{p_k})$.
Which is where I am stuck, as I don't quite see how to prove that the above product (or product squared) is greater than $\sqrt n$ (or $n$ in the product squared case). Is this the wrong angle to be approaching the problem from, or are one of my inferences altogether mistaken? 
 A: Sorry for this being very non-rigorous and not very elegant either.. as a high schooler I don't know many "high powered" tools that I could use to solve this.
Let's try to find $n$ which don't fit the statement. 
Proposition: We'll get our best bet at falsifying the statement among primorials, i.e., those numbers of the form $\displaystyle \prod_{n = 1}^k p_n$
Consider when it is not a primorial. 
Case 1: We add a $p_i$ to our primorial $p_k \#$ with $i < k$. In other words, $n$ has a repeated factor. Then $p_i \phi(n) = \phi (np_i)$. Thus, by doing this we increase the left hand side by $p_i$ while increasing our right hand side only by $\sqrt p_i$, so this is unproductive. Conclusion, we have at most of one of any prime factor in $n$.
Case 2: What if we didn't take the smallest primes. What if we changed a factor $p_i$ in $p_k \#$ into a $p_j$ with $i < k <j$? Then we would increase the left hand side by a factor of $(\frac{p_j}{p_i})(\frac{p_i}{p_i-1})(\frac{p_j-1}{p_j})$ while increasing the left hand side by a lesser factor of $\sqrt{\frac{p_j}{p_i}}$. Conclusion, we take the smallest primes we can.
Furthermore, once we find one primorial that works for the statement, all further primorials work as well. Consider if $p_i \#$ doesn't work, i.e. $\phi(p_i \#) \geq \sqrt{p_i \#}$. Then $\phi (p_{i+1} \#) = \phi(p_{i+1})\phi(p_i \#) \geq \sqrt{p_{i+1}} \sqrt{p_i \#}$, since $\phi(p_{i+1}) = p_{i+1} - 1 \geq \sqrt{p_{i+1}}$ for all $i \geq 1$
So, by inspection we find $n = 2, n = 6$ don't work. $n = 24$ works because $6 \geq \sqrt{24}$, so no further primorials work either.
To finish, we just need to do some inspection in $n =2$ and $n = 6$ to make sure that variations of them work. For $n = 2$, we check $2^2$, and find that $\phi (4) = 2 \geq 2$ works. For $n = 6$, we check $2^2 \times 3$ and $2 \times 3^2$, and find that both of them work. Thus, $n = 2, n = 6$ are the only integers $n$ which do not satisfy the statement.
