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Let $f: [0,1] \to \mathbb{R}_+$ be a decreasing and concave function.

Then $\forall a,b\in[0,1], f(\sqrt{ab})\geq \sqrt{f(a)f(b)}$.

Is it true ?

Source : les dattes à Dattier

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    Essentially, the question is whether $f(x) \ge x$ for all $x \in [0,1]$. – 2017-01-20
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    Que voulez vous dire par "under multiplicity" dans votre titre ? Sous-multiplicité ? – 2017-01-20
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    convexité multiplicative. – 2017-01-20
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    @Datier, concavité multiplicative. Geometric concavity - this name is also possible. – 2017-01-20

2 Answers 2

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Now, after the correction made by OP, this is true by AM-GM inequality. Indeed, for any $x,y\ge 0$ we have $\sqrt{xy}\le\frac{x+y}{2}$. Then $$f(\sqrt{ab})\ge f\left(\frac{a+b}{2}\right)\ge\frac{f(a)+f(b)}{2}\ge\sqrt{f(a)f(b)}.$$

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    Merci beaucoup : thank you very much. – 2017-01-20
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Try $f(x)=1-x^2$ with $a=b=0.8$.

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    This concerns the earlier version of the post. It was $\sqrt{ab}$ on the right hand side and I gave the counterexample. – 2017-01-20