mavropnevma wrote: Let $\mathcal{F}$ be the family of bijective increasing functions $f\colon [0,1] \to [0,1]$, and let $a \in (0,1)$. Determine the best constants $m_a$ and $M_a$, such that for all $f \in \mathcal{F}$ we have \[m_a \leq f(a) + f^{-1}(a) \leq M_a.\] (Dan Schwarz) 1) $M_0=0$ and $M_1=2$ and $M_a=a+1$ $\forall a\in(0,1)$ ============================= If $a=0$, then $f(0)=0$ $\forall f$ and so $M_0=0$ If $a=1$, then $f(1)=1$ $\forall f$ and so $M_1=2$ If $1>a>0$ : If $f(a)=a$, then $f(a)+f^{-1}(a)=2a\le a+1$ If $f(a)<a$, then $f(a)+f^{-1}(a)<a+1$ If $1\ge f(a)>a$, then $f^{-1}(a)<a$ and so $f(a)+f^{-1}(a)<a+1$ So $M_a\le a+1$ Choosing $f_n(x)$ (with $n$ great enough) as the three segments curve defined by the four points $(0,0)$, $(a-\frac 1n,a)$, $(a,1-\frac 1n)$ and $(1,1)$, We get $f_n(a)+f_n^{-1}(a)=1+a-\frac 2n$ and so $f(a)+f^{-1}(a)$ may be as near as we want of $a+1$ Q.E.D. 2) $m_0=0$ and $m_1=2$ and $m_a=a$ $\forall a\in(0,1)$ ============================= If $a=0$, then $f(0)=0$ $\forall f$ and so $m_0=02$ If $a=1$, then $f(1)=1$ $\forall f$ and so $m_1=2$ If $0<a<1$ : If $f(a)=a$, then $f(a)+f^{-1}(a)=2a\ge a$ If $f(a)<a$, then $f^{-1}(a)>a$ and so $f(a)+f^{-1}(a)>a$ If $f(a)>a$, then $f(a)+f^{-1}(a)>a$ So $m_a\ge a$ Choosing $f_n(x)$ (with $n$ great enough) as the three segments curve defined by the four points $(0,0)$, $(a,\frac 1n)$, $(a+\frac 1n,a)$ and $(1,1)$, We get $f_n(a)+f_n^{-1}(a)=a+\frac 2n$ and so $f(a)+f^{-1}(a)$ may be as near as we want of $a$ Q.E.D.