Show at least one of the inequality must be true

[anemone]

Let $a_1,\,a_2,\,\cdots,\,a_{12}$ be positive numbers. Show that at least of the following must be true:

$\dfrac{a_1}{a_2}+\dfrac{a_3}{a_4}+\dfrac{a_5}{a_6}+\dfrac{a_7}{a_8}+\dfrac{a_9}{a_{10}}\ge 5$,

$\dfrac{a_{11}}{a_{12}}+\dfrac{a_2}{a_1}+\dfrac{a_4}{a_3}+\dfrac{a_6}{a_5}\ge 4$, or

$\dfrac{a_8}{a_7}+\dfrac{a_{10}}{a_9}+\dfrac{a_{12}}{a_{11}}\ge 3$

 

[Albert1]

Let $a_1,\,a_2,\,\cdots,\,a_{12}$ be positive numbers. Show that at least of the following must be true:

$\dfrac{a_1}{a_2}+\dfrac{a_3}{a_4}+\dfrac{a_5}{a_6}+\dfrac{a_7}{a_8}+\dfrac{a_9}{a_{10}}\ge 5---(1)$,

$\dfrac{a_{11}}{a_{12}}+\dfrac{a_2}{a_1}+\dfrac{a_4}{a_3}+\dfrac{a_6}{a_5}\ge 4---(2)$, or

$\dfrac{a_8}{a_7}+\dfrac{a_{10}}{a_9}+\dfrac{a_{12}}{a_{11}}\ge 3---(3)$

Spoiler

if all three statements are false then:
$\dfrac{a_1}{a_2}+\dfrac{a_3}{a_4}+\dfrac{a_5}{a_6}+\dfrac{a_7}{a_8}+\dfrac{a_9}{a_{10}}< 5---(4)$,

$\dfrac{a_{11}}{a_{12}}+\dfrac{a_2}{a_1}+\dfrac{a_4}{a_3}+\dfrac{a_6}{a_5}< 4---(5)$, and

$\dfrac{a_8}{a_7}+\dfrac{a_{10}}{a_9}+\dfrac{a_{12}}{a_{11}}< 3---(6)$
by $AM-GM$ again:
it should be:
$\dfrac{a_1}{a_2}\times\dfrac{a_3}{a_4}\times\dfrac{a_5}{a_6}\times\dfrac{a_7}{a_8}\times\dfrac{a_9}{a_{10}}<1 ---(7)$,

$\dfrac{a_{11}}{a_{12}}\times\dfrac{a_2}{a_1}\times\dfrac{a_4}{a_3}\times\dfrac{a_6}{a_5}<1---(8)$, and

$\dfrac{a_8}{a_7}\times\dfrac{a_{10}}{a_9}\times\dfrac{a_{12}}{a_{11}}<1---(9)$
but:
$(7)\times(8)\times(9)=1$,it is a contradiction
so at least one of the three must be true