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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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
Gold Member
MHB
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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
Gold Member
MHB
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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