The discussion centers on proving the inequality $\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{\dfrac{1}{c}+\dfrac{1}{d}} \le \dfrac{1}{\dfrac{1}{a+c}+\dfrac{1}{b+d}}$ for positive real numbers $a, b, c, d$. A solution approach involves manipulating the terms to show that the right-hand side is always greater than zero. The calculations demonstrate that the expression simplifies to a non-negative form, confirming the inequality holds. Participants express challenges with the complexity of the fractions involved. The discussion concludes with an appreciation for the clarification provided by one member.
#1
anemone
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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 $a, b, c, d$.
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.
I can't think of another thing that solved with the limits.
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
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:
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!
#4
mente oscura
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0
anemone said:
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
anemone, would already be can it Digest better?(Poolparty)
Regards.
#5
anemone
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Thank you again menteoscura for the clarification post. I appreciate it!
#6
Albert1
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anemone said:
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 $a, b, c, d$.
$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}}---
(1)$
$ \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
Last edited:
#7
kaliprasad
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Albert said:
$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}+\dfrac{1}{ \dfrac{1}{c}+\dfrac{1}{d}}---
(1)$
$ \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