Inequalities: If (1)/(ab)+(1)/(bc)+(1)/(cd)+(1)/(da)=1, prove that...

[Math is tough]

Please see attached link for the question, as well as what I have done. Any hints on how to solve this problem will be very much appreciated. Thanks!

If (1)/(ab)+(1)/(bc)+(1)/(cd)+(1)/(da)=1, prove that:

abcd+16>=8\sqrt((a+c)((1)/(a)+(1)/(c)))+8\sqrt((b+d)((1)/(b)+(1)/(d)))

Note that a,b,c and d are all positive real numbers.

What I have done:
cd+ad+ab+bc=abcd
(d+b)(a+c)=abcd
abcd+16≥8√(abcd)
If I can prove that this is greater than the right hand side of the inequality,then I will be done? At least this is what I think...

Regarding any misunderstandings about sarcasm, I would like to clarify that I sincerely meant to appreciate any help given, with no mockery intended.Perhaps it was due to the wrong punctuation mark used that caused the misunderstandings?

 

[stapel]

Please see attached link for the question, as well as what I have done. Any hints on how to solve this problem will be very much appreciated. Thanks!

\(\displaystyle \mbox{If }\, \dfrac{1}{ab}\, +\, \dfrac{1}{bc}\, +\, \dfrac{1}{cd}\, +\, \dfrac{1}{da}\, =\, 1,\, \mbox{ prove that}\)

\(\displaystyle abcd\, +\, 16\, \geq\, 8\, \sqrt{(a\, +\, c)\, \left(\dfrac{1}{a}\, +\, \dfrac{1}{c}\right)\,}\, +\, 8\, \sqrt{(b\, +\, d)\, \left(\dfrac{1}{b}\, +\, \dfrac{1}{d}\right)\,}\)

Note that a, b, c, and d are all positive real numbers.

I'm not seeing any "attached link"...? Please reply here with the rest of the information for this exercise (what you've tried, etc). When you reply, please confirm or correct my typeset version of (what I think) you've posted. Thank you!

 

[Math is tough]

Inequality

Yes, this question is exactly what I meant!
Regarding my approach, I somehow think that it involves the use of AM-GM inequality. Currently, what I have worked out so far:

cd+ad+ab+bc=abcd
(d+b)(a+c)=abcd
abcd+16≥8√(abcd)
Maybe I am missing the point
of the question, but I think if I can prove that 8√(abcd) is greater than the right hand side of the inequality, then I will have proved the original inequality?

 

[Math is tough]

Inequalities

Thanks for helping?

 

[Deleted member 4993]

Thanks for helping?

Sarcasm will not bring any help!!

 

[Math is tough]

Inequalities

Sorry for any misunderstanding, I didn't meant for it to be sarcasm, maybe you interpreted it in the wrong way? I am sincerely appreciating any help and would like to clarify that no sarcasm was involved. Maybe the punctuation mark was used wrongly