# What are Contra Harmonic Mean and Inverse Contra Harmonic Mean? [closed]

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

## closed as unclear what you're asking by Michael Rozenberg, Lee David Chung Lin, StubbornAtom, Shogun, Martin RJun 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.