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Are they related to Inequality? Like the $$ AM \times HM = GM^2 $$

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closed as unclear what you're asking by Michael Rozenberg, Lee David Chung Lin, StubbornAtom, Shogun, Martin R Jun 11 at 4:54

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At least, both inequalities are wrong: $$\frac{a_1+a_2+...+a_n}{n}\cdot\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}\geq\left(\sqrt[n]{a_1a_2...a_n}\right)^2$$ and $$\frac{a_1+a_2+...+a_n}{n}\cdot\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}\leq\left(\sqrt[n]{a_1a_2...a_n}\right)^2.$$ Indeed, for $n=3$, $a_1=x^3$, $a_2=y^3$ and $a_3=z^3$, where $x$, $y$ and $z$ are positives we obtain: $$\frac{a_1+a_2+...+a_n}{n}\cdot\frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}}-\left(\sqrt[n]{a_1a_2...a_n}\right)^2=$$ $$=\frac{(x^3+y^3+z^3)x^3y^3z^3}{x^3y^3+x^3z^3+y^3z^3}-x^2y^2z^2=\frac{x^2y^2z^2\sum\limits_{cyc}(x^4yz-x^3y^3)}{\sum\limits_{cyc}x^3y^3}=$$ $$=\frac{x^2y^2z^2(x^2-yz)(y^2-xz)(z^2-xy)}{\sum\limits_{cyc}x^3y^3}$$ and we see that the last expression can be negative and can be positive.

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  • $\begingroup$ This is good Michael, but it's the solutions for the above inequality. This doesn't tell anything about Contra Harmonic and Inverse Contra Harmonic means. Thank you for the above proof , but was looking for information about the two means. $\endgroup$ – Vivek Adi Jun 12 at 13:18

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