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Let $f:\mathbb{R} \to \mathbb{R}$ be a $C^1$ function, define $F:(x,y) \mapsto(x+f(y),y+f(x))$

I need to prove that if there exists a constant $0

$F'(z;v).v \geq b|v|^2$ for every $z,v\in \mathbb{R^2} $

$F'(z;v)$ denotes the directional derivative at z in the direction of v

I know that $F'(z;v).v=v_1^2+v_2^2+2v_1v_2 \frac{f'(x)+f'(y)}{2}$, but from here I don't know how I could obtain the correct inequality, everything I can think of gives me $F'(z;v).v \leq something$.

Any advices?

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    $F : \mathbb R^2 \to \mathbb R^2$. I'm unclear on what you mean by $F'$. – 2017-02-08
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    The directional derivative at $z$ in the direction of $v$ – 2017-02-08
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    I see. Thanks for clarifying. – 2017-02-08

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Note that $$\left| 2v_1 v_2 \left( \frac{f'(x) + f'(y)}{2} \right) \right| \le 2a |v_1| |v_2|$$ by hypothesis and $$2a |v_1| |v_2| \le a(v_1^2 + v_2^2)$$ in light of the elementary inequality $xy \le \frac{x^2 + y^2}{2}$ whenever $x,y \ge 0$. You end up with, in turn, $$\left| 2v_1 v_2 \left( \frac{f'(x) + f'(y)}{2} \right) \right| \le a(v_1^2 + v_2^2),$$ $$ 2v_1 v_2 \left( \frac{f'(x) + f'(y)}{2} \right)\ge - a(v_1^2 + v_2^2),$$ $$ v_1^2 + v_2^2 + 2v_1 v_2 \left( \frac{f'(x) + f'(y)}{2} \right)\ge (1- a)(v_1^2 + v_2^2),$$ $$ F'(x,y;v) \cdot v \ge (1-a)|v|^2.$$