<s>[INST] 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.
 [/INST] Subtract the inequality from the equation
$$
1=(a+b)^2=a^2+2ab+b^2\\
0\le(a-b)^2=a^2-2ab+b^2
$$
to get
$$
1\ge4ab\implies\frac14\ge ab\tag{1}
$$
Apply $(1)$ to get
$$
\begin{align}
\left(1+\frac1a\right)\left(1+\frac1b\right)
&=1+\frac1a+\frac1b+\frac1{ab}\\
&=1+\frac{a+b}{ab}+\frac1{ab}\\[3pt]
&=1+\frac2{ab}\\[3pt]
&\ge1+\frac2{1/4}\\[8pt]
&=9\tag{2}
\end{align}
$$
</s>