Prove Inequality: $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}$

[anemone]

Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.

 

[mente oscura]

Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.

Hello.

Spoiler

I can't think of another thing that solved with the limits.

Making calculations.(Whew)

T=\dfrac{1}{1+\frac{d}{b}+\frac{b}{a}+\frac{d}{a}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}+ \dfrac{1}{1+\frac{a}{b}+\frac{c}{b}+\frac{c}{a}}+ \dfrac{1}{1+\frac{a}{d}+\frac{a}{c}+\frac{c}{d}} \le{1}

We note that, when they are the same two by two, the result is 1.Now let's look:

\displaystyle\lim_{a \to{+}\infty}{T}=\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}

\displaystyle\lim_{b \to{+}\infty}{} (\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}})=1

It would be like if we choose any of the other variables, because "T" combinations are symmetric.

Regards.

 

[anemone]

Hello.

Spoiler

I can't think of another thing that solved with the limits.

Making calculations.(Whew)

T=\dfrac{1}{1+\frac{d}{b}+\frac{b}{a}+\frac{d}{a}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}+ \dfrac{1}{1+\frac{a}{b}+\frac{c}{b}+\frac{c}{a}}+ \dfrac{1}{1+\frac{a}{d}+\frac{a}{c}+\frac{c}{d}} \le{1}

We note that, when they are the same two by two, the result is 1.Now let's look:

\displaystyle\lim_{a \to{+}\infty}{T}=\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}

\displaystyle\lim_{b \to{+}\infty}{} (\dfrac{1}{1+\frac{d}{b}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}})=1

It would be like if we choose any of the other variables, because "T" combinations are symmetric.

Regards.

Thanks for your post, mente oscura! I have been trying to digest what you have come up with for some time, hoping to see the light at the end of the tunnel but I am sorry, I couldn't possibly continue from there.

I want to share the solution that I found with you and the rest of the members:

If we shift everything from the left to the right, we get

$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$

$0 \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}-\left( \dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \right)$

$\dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}-\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \right) \ge 0$

That is to say, if we can prove the RHS is always greater than zero for all positive real numbers $a, b, c, d$, then we are done. Let's see how far this will take us:

$\begin{align*}\small \dfrac{1}{ \dfrac{1}{a+c}+ \dfrac{1}{b+d}}-\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \right)&=\dfrac{1}{\dfrac{b+d+a+c}{(a+c)(b+d)}}-\left(\dfrac{1}{\dfrac{b+a}{ab}}+\dfrac{1}{\dfrac{d+c}{cd}} \right)\\&=\dfrac{(a+c)(b+d)}{a+b+c+d}-\dfrac{ab}{a+b}-\dfrac{cd}{ c+d}\\&=\\&= \small\dfrac{(a+c)(b+d)(a+b)(c+d)-ab(c+d)(a+b+c+d)-cd(a+b)(a+b+c+d)}{(a+b+c+d)(a+b)(c+d)}\\&=\tiny \dfrac{(a+c)(b+d)(a+b)(c+d)-ab(a+b)(c+d)-ab(c+d)^2-cd(a+b)^2-cd(a+b)(c+d)}{(a+b+c+d)(a+b)(c+d)}\\&= \dfrac{(a+b)(c+d)((a+c)(b+d)-ab-cd)-cd(a+b)^2-ab(c+d)^2}{(a+b+c+d)(a+b)(c+d)}\\&=\small\dfrac{(a+b)(c+d)(ab+ad+bc+cd-ab-cd)-cd(a^2+2ab++b^2)-\tiny ab(c^2+2cd++d^2)}{(a+b+c+d)(a+b)(c+d)}\\&= \dfrac{(ac+ad+bc+bd)(ad+bc)-a^2cd-2abcd-b^2cd-abc^2-2abcd-abd^2}{(a+b+c+d)(a+b)(c+d)}\\&=\small\dfrac{a^2cd+abc^2+a^2d^2+abcd+abcd+b^2c^2+abd^2+b^2cd-a^2cd-2abcd-b^2cd-abc^2-2abcd-abd^2}{(a+b+c+d)(a+b)(c+d)}\\&=\dfrac{a^2d^2-2abcd+b^2c^2}{(a+b+c+d)(a+b)(c+d)}\\&=\dfrac{(ad-bc)^2}{(a+b+c+d)(a+b)(c+d)}\\& \ge 0\end{align*}$

since $(ad-bc)^2 \ge 0$ and $(a+b+c+d)(a+b)(c+d) >0$ for all positive real numbers $a, b, c, d$ and we're now done.

I am getting so tired of keep previewing the post because of so many fractions and terms that I have to deal with in this particular problem...My head is hurting me so bad now and my vision is blurred for a moment!

 

[mente oscura]

Thanks for your post, mente oscura! I have been trying to digest what you have come up with for some time, hoping to see the light at the end of the tunnel but I am sorry, I couldn't possibly continue from there.

I want to share the solution that I found with you and the rest of the members:

If we shift everything from the left to the right, we get

$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$

...

Hello. Uff.:p

My calculations are:

\dfrac{1}{\frac{1}{a}+\frac{1}{b}}+\dfrac{1}{\frac{1}{c}+\frac{1}{d}} \le \dfrac{1}{\frac{1}{a+c}+\frac{1}{b+d}}\dfrac{1}{\frac{a+b}{ab}}+\dfrac{1}{\frac{c+d}{cd}} \le \dfrac{1}{\frac{a+b+c+d}{(a+c)(b+d)}}\dfrac{\frac{c+d}{cd}+\frac{a+b}{ab}}{\frac{(a+b)(c+d)}{abcd}} \le \dfrac{(a+c)(b+d)}{a+b+c+d}\dfrac{ab(c+d)+cd(a+b)}{(a+b)(c+d)} \le \dfrac{(a+c)(b+d)}{a+b+c+d}\dfrac{[ab(c+d)+cd(a+b)](a+b+c+d)}{(a+b)(c+d)(a+c)(b+d)} \le{1}\dfrac{[ab(c+d)+cd(a+b)] \cancel{(a+c)}}{(a+b)(c+d) \cancel{(a+c)}(b+d)}+\dfrac{[ab(c+d)+cd(a+b)] \cancel{(b+d)}}{(a+b)(c+d)(a+c) \cancel{(b+d)}} \le{1}\dfrac{ab \cancel{c+d}}{(a+b) \cancel{(c+d)}(b+d)}+\dfrac{cd \cancel{a+b}}{\cancel{(a+b)}(c+d)(b+d)}+\dfrac{ab \cancel{c+d}}{(a+b) \cancel{(c+d)}(a+c)}+\dfrac{cd \cancel{a+b}}{\cancel{(a+b)}(c+d)(a+c)} \le{1}\dfrac{ab}{ab+ad+b^2+bd}+\dfrac{cd}{bc+bd+cd+d^2}+\dfrac{ab}{a^2+ab+ac+bc}+\dfrac{cd}{ac+ad+c^2+cd} \le{1}

The end:

\dfrac{1}{1+\frac{d}{b}+\frac{b}{a}+\frac{d}{a}}+ \dfrac{1}{1+\frac{b}{d}+\frac{b}{c}+\frac{d}{c}}+ \dfrac{1}{1+\frac{a}{b}+\frac{c}{b}+\frac{c}{a}}+ \dfrac{1}{1+\frac{a}{d}+\frac{a}{c}+\frac{c}{d}} \le{1}

And, from here, as set out in my previous post.

anemone, would already be can it Digest better?(Poolparty)

Regards.

 

[anemone]

Thank you again mente oscura for the clarification post. I appreciate it!

 

[Albert1]

Prove that $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for all positive real numbers $a, b, c, d$.

Spoiler

$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}}---
(1)$
$ \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}----(2)$
using $AP\geq GP :$
(1)$\leq\dfrac{\sqrt {ab}}{2}+\leq\dfrac{\sqrt {cd}}{2} $---(3)
(2)$\leq\dfrac{\sqrt {(a+c)(b+d)}}{2}$---(4)
$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+cd}{4}----(6)$
again using $AP\geq GP :$
$(6)\geq (5)$
and the proof is done

 

[kaliprasad]

Spoiler

$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}}---
(1)$
$ \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}----(2)$
using $AP\geq GP :$
(1)$\leq\dfrac{\sqrt {ab}}{2}+\leq\dfrac{\sqrt {cd}}{2} $---(3)
(2)$\leq\dfrac{\sqrt {(a+c)(b+d)}}{2}$---(4)
$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+bd}{4}----(6)$
again using $AP\geq GP :$
$(6)\geq (5)$
and the proof is done

may be I am missing something

Spoiler

but a< b, c < d, b < d does not imply a < c

 

[Albert1]

sory a typo in (6)

Spoiler

$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+cd}{4}----(6)$
$ad+bc\geq 2\sqrt {abcd}$

 

[kaliprasad]

sory a typo in (6)

Spoiler

$(3)^2=\dfrac {ab+cd+2\sqrt {abcd}}{4} ----(5)$
$(4)^2=\dfrac {ab+ad+bc+cd}{4}----(6)$
$ad+bc\geq 2\sqrt {abcd}$

Spoiler

(6) > (5) => (3) < (4) is true

But (3) < (4) and (1) < (3) and (2) < (4) does not mean (1) < (2) .

 

[Albert1]

Spoiler

(6) > (5) => (3) < (4) is true

But (3) < (4) and (1) < (3) and (2) < (4) does not mean (1) < (2) .

Spoiler

if (1)=(3) then a=b and c=d---(7)
if (2)=(4) then a+c=b+d---(8)
both (7) and (8) must hold together
(here a,b,c,d>0)