Prove inequality and when does equality hold? Prove that for any real positive numbers $a$ and $b$
$$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq 4.$$
When does equality hold?
 A: Hint:
$$ \;\; \text{For any two real numbers $a$ and $b$, } \;\;(ab - 1)^2 \ge 0 $$
Simplify the above expression and try to use the positivity of $a$ and $b$ to arrive at your inequality. And if you do get your inequality then the answer to your follow-up question is pretty much embedded in the hint.  
A: By AM-GM inequality,  
$\bigg(a+\dfrac{1}{b}\bigg)\ge2\sqrt{\dfrac{a}{b}}$  
Again,  
$\bigg(b+\dfrac{1}{a}\bigg)\ge2\sqrt{\dfrac{b}{a}}$  
By multiplying the above equations,  
$\bigg(a+\dfrac{1}{b}\bigg)\cdot\bigg(b+\dfrac{1}{a}\bigg)\ge 2\cdot2\cdot \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{a}{b}}=4$  
Equality occurs only when $a=\dfrac{1}{b}$
A: $$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq 4 = ab+\frac{1}{ab} + 2.$$
Now, all you need is to see what the minimum of the value $ab+\frac{1}{ab}$ is. Since any positive real number $x$ can be written as $ab$, this is equivalent to calculating the minimum of the function $$f(x) = x+\frac1x.$$ It is easy to show that for any $x>0$, the value $$x+\frac1x$$ is greater or equal to $2$:
$$x+\frac1x \geq 2 \iff\\ x^2-2x+1\geq 0.$$
A: Let a,b two real numbers such that $x=ab$ is positive (both a,b have the same sign)
Than $x+\frac {1} {x} \geq 2$ is true for every $x>0$. Why? 
Denote $f(x)=x+\frac {1} {x} -2$
Now,
$f'(x)=1-\frac {1} {x^2} $
$f'(x)>0$ for every $x>1$
$f(1)=1+1-2=0$
therefore $f(x)>f(1)=0$ for every $x>1$
Which means  $x+\frac {1} {x} \geq 2$
I will leave you to prove the inequality for $0<x<1$.
A: Use the AM-GM inequality to obtain
$$
\frac{\bigl(a+\frac1b\bigr)\bigl(b+\frac1a\bigr)}4=\frac{ab+1+1+\frac1{ab}}{4}\ge1.
$$
Equality holds when $ab=1$.
A: Let's just multiply it out.
$ab + \frac{1}{ab} \ge 2$
Intuïtively we can see that the left expression has minimal value when a=b=1. Its value then is 4.
