prove $\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le\frac{1}{64abcd}$

Question

Let $a,b,c,d$ be positive real no such that $a+b+c+d\le1$

prove that$$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}\le \frac{1}{64abcd}$$

I did this way$$\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}\le\frac{1}{64abcd}$$

$\implies$ $a^2cd+b^2cd+c^2ab+d^2ab\le\frac{1}{64}$

$\implies$ $(ac+bd)(ad+bc)\le\frac{1}{64}$

now I don't know how to proceed.

Answer 1

By AM-GM $(ac+bd)(ad+bc)\leq\left(\frac{ac+bd+ad+bc}{2}\right)^2=\left(\frac{(a+b)(c+d)}{2}\right)^2\leq\left(\frac{\left(\frac{a+b+c+d}{2}\right)^2}{2}\right)^2\leq\frac{1}{64}$

Answer 2

$\frac{a}{b}+\frac{b}{a}+\frac{c}{d}+\frac{d}{c}=\frac{a^2cd+b^2cd+c^2ab+d^2ab}{abcd}$

$\implies(ac+bd)(ad+bc)\le\frac{1}{64}$

Now $(ac+bd)(ad+bc)\le\frac{ac+bd+ad+bc}{4}$

$=\frac{(a+b)(c+d)}{4}\le\frac{(a+b+c+d)}{64}$

As $a+b+c+d\le1$

$\implies\frac{(a+b+c+d)^4}{64}\le\frac{1}{64}$

Hence proved