Prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without limits I'm facing a problem where I need to prove that $(1+a^2)(1+b^2)\geq 4ab$ for all $a,b>0$ without any use of limits.
 I have tried 2 different methods that I'm not really sure about:
1) Taking out factors $a$ and $b$ so I get: $$ab(a+\frac{1}{a})(b+\frac{1}{b})\geq 4ab$$ $$(a+\frac{1}{a})(b+\frac{1}{b})\geq 4$$
and then split for cases: suppose $a,b<0.5$ so both $\frac{1}{a}>2$ and $\frac{1}{b}>2$ 
Therefore:$(a+\frac{1}{a})$>2 and $(b+\frac{1}{b})$>2 so $(a+\frac{1}{a})(b+\frac{1}{b}) >4$ and so on.
In cases both are smaller but very close to $1$ is there any general way to prove that $(a+\frac{1}{a})>2$?
2) The other method I've tried was more direct and algebric, but wasn't too helpfull:
$$(1+a^2)(1+b^2)=1+a^2+b^2+(ab)^2=1+(a-b)^2+2ab+(ab)^2$$
So I tried:
$$(1+a^2)(1+b^2)\geq4ab \rightarrow 1+(a-b)^2+(ab)^2\geq2ab$$
but without limits it's too hard to say anything about cases where $a\rightarrow0$ or $b\rightarrow0$ or both.
If there's anything I'm missing I'd like to be enlightened.

Writing
$$1+a^2+b^2+a^2b^2\geq 4ab$$ and this is
$$a^2+b^2-2ab+1+a^2b^2-2ab\geq 0$$ and this is $$(a-b)^2+(ab-1)^2\geq 0$$
Or $$ab+\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}\geq 2+2=4$$
since $$x+\frac{1}{x}\geq 2$$ if $$x\geq 0$$
