Prove inequality and when does equality hold?

Question

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?

Answer 1

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.

Answer 2

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 $a=\dfrac{1}{b}$

Answer 3

$$\left(a+\frac{1}{b}\right)\left(b+\frac{1}{a}\right) \geq 4 = ab+\frac{1}{ab} + 2.$$

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 $2$: $$x+\frac1x \geq 2 \iff\\ x^2-2x+1\geq 0.$$

Answer 4

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 $0<x<1$.

Answer 5

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 $ab=1$.

Answer 6

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.