Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$

Question

Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$

My attempt:

$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\ & =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\ &=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\ &=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\ &=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$

I don't know how to continue from this.

Can someone help me?

Answer 1

Use AM-GM. $\frac{ad + bc}{2} \ge \sqrt{abcd}$. Squaring both sides, you get the answer. A tiny tip: If everything is positive, and you have an inequality, think about AM-GM once at least.

Answer 2

Now, $$\frac{(ad+bc)^2}{abcd}-4=\frac{a^2d^2-2abcd+b^2c^2}{abcd}=\frac{(ad-bc)^2}{abcd}\geq0.$$Also, by C-S $$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq\left(\sqrt{\frac{b}{a}\cdot\frac{a}{b}}+\sqrt{\frac{d}{c}\cdot\frac{c}{d}}\right)^2`=4$$

Answer 3

Taking a different approach entirely, note that

$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)=1+{ad\over bc}+{bc\over ad}+1$$

Thus, letting $ad/bc=x$ and noting that $x\gt0$, we see that

$$\left({b\over a}+{d\over c}\right)\left({a\over b}+{c\over d}\right)\ge4\iff x+{1\over x}\ge2\iff x^2-2x+1\ge0\iff(x-1)^2\ge0$$

(Note, the stipulation $x\gt0$ is important is multiplying both sides of the inequality $x+1/x\ge2$ by $x$ to get to $x^2+1\ge2x$.)

Answer 4

To continue from

$\tag 1 \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) = \dfrac{(ad+bc)^2}{abcd}$

set

$\quad u = ad$

and

$\quad v = bc$

Then substituting in the rhs of $\text{(1)}$, we have

$\quad \dfrac{(u+v)^2}{uv} \ge 4 \text{ iff } (u-v)^2 \ge 0$

Note that if we let $a,b,c,d \in \Bbb R$ satisfy $abcd \gt 0$ then

$\quad \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \ge 4$

and if $abcd \lt 0$ then

$\quad \left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right) \le 4$

Answer 5

Let $x=a/b$, $y=c/d$, you'll get $2+x/y+y/x$. Now use that for any positive number the sum of that number and its reciprocal is at least $2$.