Proving the equivalence of two inequalities

Question

Let $f:\mathbb{R} \to \mathbb{R}$ be a function that admits primitives on $\mathbb{R}$ and $F:\mathbb{R} \to \mathbb{R}$ one primitive (i.e. $F'(x)=f(x), \forall x \in \mathbb{R}$).
Prove that the following two inequalities are equivalent: \begin{align} a)&& (f(x)-f(y))(x-y) &\ge (x-y)^2, && \forall x, y \in \mathbb{R} \\b)&& F(x)-F(y) &\ge \frac{1}{2}(x-y)^2+f(y)(x-y), &&\forall x, y \in \mathbb{R} \end{align} My approach was to prove $b) \implies a)$. It is obvious, because we if $b)$ is true for all pairs $(x, y)$ then it is true for $(y, x)$ by adding the two relations we obtain "$a)$". But I have not found a way to prove the converse statement. My first thought was to use the mean value theorem to get $F$ into the equation, but the mean value theorem implies the existence of a point, which would not satisfy the $\forall$ quantifier.

Answer 1

For any primitive $F$, $F(X) - F(Y) = \int_Y^X f(x) dx$. In the case $x>y$, ineq. $a)$ becomes

$$f(x) - f(y) \ge x-y$$

Let $y \le Y \le x \le X $. If we integrate this inequality in $x$ from $Y$ to $X$, we get $$ F(X)-F(Y) - (X-Y)f(y) \ge \frac12(X^2 - Y^2) - y(X-Y) \tag{*} $$

Making the choice $y=Y$ and $x=X$ we obtain $$ F(x) -F(y) \ge \frac12(x^2 + y^2) - xy + (x-y)f(y) = \frac12(x-y)^2 + (x-y)f(y) $$ Which is $b)$.

So we are done with $a)\implies b)$, for when $x>y$. The case $x=y$ is trivial. For the case $x<y$, ineq. $a)$ instead becomes $$ f(y)-f(x) \ge y-x$$

Integrating now in $x$ from $X$ to $Y$ (this time $X\le x\le Y\le y$) we get $$ (Y-X)f(y)-(F(Y)-F(X))\ge y(Y-X) -\frac12(Y^2-X^2)$$

which is exactly inequality $(\text *)$ after rearranging, so setting $y=Y$ and $x=X$ again gives the result.