<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] If the inequality$$\frac{a^2+b^2}{2}\ge\left(\frac{a+b}{2}\right)^2$$is used, then we can deduce that $a^2+b^2\ge2ab$ must hold.  This can be seen from squaring the terms on the LHS and simplifying the resulting expression.
Now, $$ab\le\frac12(a^2+b^2)=\frac{(a+b)^2-2ab}{2}=\frac12-ab$$This implies that $$2ab\le\frac12$$which in turn gives the first inequality$$ab\le\frac14$$
For the second inequality, we have
$$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)=1+\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=1+\frac{a+b+1}{ab}=1+\frac{2}{ab}$$But we just showed that $ab\le\frac14$ which means that $\frac{1}{ab}\ge 4$.  Using this, we find $$\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\ge1+(2)(4)=9$$and that completes the proof for the second inequality!
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