Prove that $ab \leq \frac14$ and $ (1+1/a)(1+1/b)\ge 9$ when $a+b=1, a \ge 0, b \ge 0$ Our teacher gave us some identities firstly and said we could use one of them to prove it. The identities are: $$\frac{(a^2+b^2)}{2}≥\left(\frac{(a+b)}{2}\right)^2$$ $$(x+y)^2≥2xy$$ and $$\frac{(x+y)}{2} \ge \sqrt{xy}$$ 
Here's what I did to solve this exercise:
$$\frac{a+b}2 \ge \sqrt{ab} \implies \frac12 \ge \sqrt{ab} \implies \frac14\ge ab \implies ab\le\frac14$$
$$\left(1+\frac1a\right)\left(1+\frac1b\right) = 1+\frac1b+\frac1a+\frac1{ab} \ge 9 \Rightarrow \frac{a+b+1}{ab}\ge8 \Rightarrow \frac2{ab}\ge8 \Rightarrow 2\le8ab \Rightarrow ab\le\frac14$$
I came to a known point, but I'm not sure that this is the right form. If there is any other more clear proof please show it to me.

If you're only allowed to use $x+y\geq 2\sqrt{xy}$ (for $x,y\geq 0$), then
$$
1=a+b\geq2\sqrt{ab}\implies ab\leq\frac{1}{4}
$$
and
$$
(1+1/a)(1+1/b)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}\geq 1+\frac{2}{\sqrt{ab}}+\frac{1}{ab}\geq1+\frac{2}{\sqrt{1/4}}+\frac{1}{1/4}=9.
$$
