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improved formatting
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edit approved May 5, 2017 at 16:26
the_fox
edit approved May 5, 2017 at 16:26
the_fox
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For a continuous function f$f: R+ -> R+\mathbb{R}^{+}\to\mathbb{R}^{+}$ does $(f(AMx)-f(GMy)  )  (f(min\frac{x+y}{2})  -f f(max\sqrt{xy}))=0 means f=0$ imply that $f$ is constant?

GivenSuppose that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true : $$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0$$ .$$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0.$$ Is it true that the only solution to this is the constant function ?

For continuous function f: R+ -> R+ does (f(AM)-f(GM)  )(f(min)-f(max))=0 means f is constant?

Given that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true : $$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0$$ . Is it true that the only solution to this is the constant function ?

For a continuous function $f:\mathbb{R}^{+}\to\mathbb{R}^{+}$ does $(f(x)-f(y))  (f(\frac{x+y}{2})  - f(\sqrt{xy}))=0$ imply that $f$ is constant?

Suppose that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true : $$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0.$$ Is it true that the only solution to this is the constant function ?

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Aditya Guha Roy
Aditya Guha Roy
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For continuous function f: R+ -> R+ does (f(AM)-f(GM) )(f(min)-f(max))=0 means f is constant?

Given that $f: \mathbb{R}^+ \to \mathbb{R}^+$ is a continuous function such that for all positive real numbers $x,y$ the following is true : $$(f(x)-f(y)) \left ( f \left ( \frac{x+y}{2} \right ) - f ( \sqrt{xy} ) \right )=0$$ . Is it true that the only solution to this is the constant function ?