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I have a proof of a complicated inequality in my book, which first does a little manipulation and after this step- $$ (1+3x) \left(1+{8y\over x} \right) \left( 1+{9z\over y}\right) \left(1+{6\over z} \right) \geq 7^4 $$ It mentions the proof is finished by Huygens Inequality.
My question is that what is Huygens Inequality? I found documents related to trigonometry while searching for this inequality online, but unfit for the proof of this step.

Please explain me what this inequality is. Thanks!

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  • $\begingroup$ First Google search result: artofproblemsolving.com/community/c1642h1004522 $\endgroup$
    – Martin R
    Commented Aug 8, 2020 at 11:25
  • $\begingroup$ @MartinR, I know, I saw. But someone commented that it is wrongly stated, so I am dubious. $\endgroup$ Commented Aug 8, 2020 at 11:31
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    $\begingroup$ You asked “what is Huygens Inequality?”.– If you already know what it is and have a specific question about it then please update your question accordingly. $\endgroup$
    – Martin R
    Commented Aug 8, 2020 at 11:43
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    $\begingroup$ @MartinR I said in the previous comment that I'm uncertain for its validity. I want an undoubted result. $\endgroup$ Commented Aug 8, 2020 at 11:53

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The Huygens's inequality it's just privet case of the Holder's inequality for $n$ sequences of the length two.

I think, it's better to use Holder in any case:

By Holder $$(1+3x) \left(1+{8y\over x} \right) \left( 1+{9z\over y}\right) \left(1+{6\over z} \right)\geq\left(1+\sqrt[4]{3x\cdot\frac{8y}{x}\cdot\frac{9z}{y}\cdot\frac{6}{z}} \right)^4=2401.$$

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    $\begingroup$ Thank you very much! I got my answer. I see I got better at inequalities. It wasn't possible without your support! $\endgroup$ Commented Aug 8, 2020 at 12:07

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