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