Assume that $F(e^{it})=e^{if(t)}$ is a diffeomorphism of the unit circle onto itself and let $A=\left|\int\_0^{2\pi}(1-F^2)\,dt\right|$ and $B=\left|\int\_0^{2\pi} F^2(1-F^2) \,dt\right|$. It seems that $A>B$ but so far I cannot prove. 


Thanks



An integral inequality for diffeomorphisms
It looks false. Note that $1-\cos x<1+\cos x-2\cos^2 x=\cos x-\cos 2x$ for $|x|\in (0,\pi/2)$. Therefore if say $|2f(t)|\in (\pi/10,\pi/3)$ for almost all $t$, the real part of $\int F^2(t)(1-F^2(t))$ is greater than that of $\int 1-F^2(t)$. If also $2f(t+\pi)=-2f(t)$ for $t\in [0,\pi]$, the imaginary part of $\int 1-F^2$ is zero due to symmetry.


