MHB Is There an Inequality Challenge with Real Numbers?

[anemone]

Let $a,\,b,\,c,\,d$ be real numbers such that $abcd=1$ and $a+b+c+d>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}$.

Prove that $a+b+c+d<\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}$.

 

[Albert1]

Let $a,\,b,\,c,\,d$ be real numbers such that $abcd=1$ and $a+b+c+d---(1)>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}---(2)$.

Prove that $a+b+c+d<\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}---(3)$.

Spoiler

let $a>b>c>d,\,\, and ,\,\,abcd=1$
we have:$(1)=a+b+c+d>(2)=a^2cd+ab^2d+abc^2+bcd^2$
if $(1)<(3)=a^2bc+b^2cd+ac^2d+abd^2$ is true, then
$-1\times(1)=-a-b-c-d>-1\times(3)$
$=-a^2bc-b^2cd-ac^2d-abd^2---(4)$
now we ony have to prove $(2)+(4)<0$
after simplification we have:
$(2)+(4)=(a-c)(b-d)(bd-ac)<0$
and the proof is done

 

[kaliprasad]

Spoiler

let $a>b>c>d,\,\, and ,\,\,abcd=1$
we have:$(1)=a+b+c+d>(2)=a^2cd+ab^2d+abc^2+bcd^2$
if $(1)<(3)=a^2bc+b^2cd+ac^2d+abd^2$ is true, then
if $-1\times(1)=-a-b-c-d>-1\times(3)$
$=-a^2bc-b^2cd-ac^2d-abd^2---(4)$
now we ony have to prove $(2)+(4)<0$
after simplification we have:
$(2)+(4)=(a-c)(b-d)(bd-ac)<0$
and the proof is done

There are a couple of issues

Spoiler

issue 1:because of non symmetry

what if a > c> b> d.

issue 2
we have shown that
$\dfrac{a}{b} + \dfrac{b}{c}+ \dfrac{c}{d}+\dfrac{d}{a} \le \dfrac{b}{a} + \dfrac{c}{b}+ \dfrac{d}{c}+\dfrac{a}{d} $
this dos not mean that
$a + b + c + d \le \dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{d}{c}+\dfrac{a}{d} $

 

[Albert1]

There are a couple of issues

Spoiler

issue 1:because of non symmetry

what if a > c> b> d.

issue 2
we have shown that
$\dfrac{a}{b} + \dfrac{b}{c}+ \dfrac{c}{d}+\dfrac{d}{a} \le \dfrac{b}{a} + \dfrac{c}{b}+ \dfrac{d}{c}+\dfrac{a}{d} $
this dos not mean that
$a + b + c + d \le \dfrac{b}{a}+ \dfrac{c}{b}+ \dfrac{d}{c}+\dfrac{a}{d} $

a>c>b>d
(a-c)(b-d)(bd-ac)<0 also is true
x>y----(m)
-x>-z---(n)
if (m)(n) both true then
(m)+(n)
0>y-z
y-z=y+(-z)must <0
so x<z

 

[anemone]

Hi Albert,

I don't think I am in the right position to judge your solution as I don't quite follow some of the arguments, but I want to share with you, kaliprasad and others of the solution that I saw online::)

Spoiler

By using the AM-GM inequality to account for the terms $\dfrac{a}{b},\,\dfrac{a}{b},\,\dfrac{b}{c},\,\dfrac{a}{d}$, we have

$\dfrac{\dfrac{a}{b}+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{a}{d}}{4}\ge \sqrt[4]{\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{a}{d}}=\sqrt[4]{\dfrac{a^4}{abcd}}=a$

Analogously,
$\dfrac{\dfrac{b}{c}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{b}{a}}{4}\ge b$,

$\dfrac{\dfrac{c}{d}+\dfrac{c}{d}+\dfrac{d}{a}+\dfrac{c}{b}}{4}\ge c$,

$\dfrac{\dfrac{d}{a}+\dfrac{d}{a}+\dfrac{a}{b}+\dfrac{d}{c}}{4}\ge d$,

Summming up all these four inequalities gives

$\dfrac{3}{4}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\right)+\dfrac{1}{4}\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}\right)\ge a+b+c+d$

In particular,

if $a+b+c+d>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}$, then $a+b+c+d<\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}$ and we are done.

 

[Albert1]

Hi Albert,

I don't think I am in the right position to judge your solution as I don't quite follow some of the arguments, but I want to share with you, kaliprasad and others of the solution that I saw online::)

Spoiler

By using the AM-GM inequality to account for the terms $\dfrac{a}{b},\,\dfrac{a}{b},\,\dfrac{b}{c},\,\dfrac{a}{d}$, we have

$\dfrac{\dfrac{a}{b}+\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{a}{d}}{4}\ge \sqrt[4]{\dfrac{a}{b}\cdot\dfrac{a}{b}\cdot\dfrac{b}{c}\cdot\dfrac{a}{d}}=\sqrt[4]{\dfrac{a^4}{abcd}}=a$

Analogously,
$\dfrac{\dfrac{b}{c}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{b}{a}}{4}\ge b$,

$\dfrac{\dfrac{c}{d}+\dfrac{c}{d}+\dfrac{d}{a}+\dfrac{c}{b}}{4}\ge c$,

$\dfrac{\dfrac{d}{a}+\dfrac{d}{a}+\dfrac{a}{b}+\dfrac{d}{c}}{4}\ge d$,

Summming up all these four inequalities gives

$\dfrac{3}{4}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\right)+\dfrac{1}{4}\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}\right)\ge a+b+c+d$

In particular,

if $a+b+c+d>\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}$, then $a+b+c+d<\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{d}{c}+\dfrac{a}{d}$ and we are done.

you did not say a,b,c,d>0
How can you solve it
"By using the AM-GM inequality "?
exam:
$-1=\dfrac{\dfrac{-1}{1}+\dfrac{-1}{1}+\dfrac{-1}{1}+\dfrac{-1}{1}}{4}\ge \sqrt[4]{\dfrac{-1}{1}\cdot\dfrac{-1}{1}\cdot\dfrac{-1}{1}\cdot\dfrac{-1}{1}}=\sqrt[4]{\dfrac{(-1)^4}{1}}=1$
does not hold

 

[anemone]

You're right, Albert(Yes)...has to check with the original problem and post back if I find it.