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I want submit to you an inequality .To beginn I know that we have this for a,b belonging to $[0;1]$ : $$(\cos(ab))^2\geq \cos(a²)\cos(b²)$$

My idea was to create a new inequality with this : $$\sqrt{\frac{\cos(\cos(ab))}{\cos(1)}}\geq 1 \geq \cos(a²)\cos(b²)$$ For any positive a and b. But it's true for a,b belonging to $[0;1]$ because we have : $$\sqrt{\frac{\cos(\cos(ab))}{\cos(1)}}\geq (\cos(ab))²\geq \cos(a²)\cos(b²)$$ And it's clear that : $$\sqrt{\frac{\cos(\cos(x))}{\cos(1)}}\geq (\cos(x))²$$ But I have no idea to generalize this for any a and b... Edit it's trivial in fact . I suggest to you an another inequality : Let $a$ and $b$ belonging to [0;1] and with the following condition $b^2\geq ab \geq a^2$ we have : $$\sqrt{cos(b^2)^{cos(a^2)}}\leq \frac{cos(cos(ab))}{cos(1)}$$ We have also for $a$ and $b$ belonging to $[0;\frac{\pi}{17}]$ : $$\sqrt{cos(b^2)^{cos(a^2)}}\leq cos(\frac{cos((ab)^2)}{cos(1)})+\frac{7\pi}{17}$$

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    In your first inequality, on the left hand side, do you mean $\cos(a^2 b^2)$ or $\cos^2(ab)$. It looks to me you mean $\cos\big((ab)^2\big) = \cos(a^2b^2)$, but I thought it best to clarify, first. – 2017-02-05
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    In the same vein, when you write, at the end, $\cos(x)²,$ do you mean $(\cos(x))^2,$ (which means $\cos^2 x$ and not $\cos(\cos x))\,$ or $\cos(x^2)$? – 2017-02-05
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    Just as an aside: The standard way of writing, e.g., $(\cos(ab))^2$ is simply $\cos^2(ab)$ which means (\cos(ab))^2 = (\cos(ab)\cdot \cos(ab)$. Regardless, you have made your question much clearer! – 2017-02-05
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    @AmWhy have you some elements to proved this last inequality ? I have a proof but I would like to know a easy proof from somebody else . – 2017-02-06
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    In the last inequality, it seems $b^2\geq ab \geq a^2$ is not necessary. – 2017-02-06

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