Let $f, g$ be functions differentiable at $1$ and such that $f(x)+x ≤ g(x)+x^2$ and $f(1) = g(1).$ Show that $f'(1)+1 = g' (1)$

Question

Let $f, g$ be functions differentiable at $1$ and such that $f(x)+x ≤ g(x)+x^2, ~x\in\Bbb R$ and $f(1) = g(1).$ Show that $f ′ (1)+1 = g ′ (1).$

When trying to do this exercise I started stating the mean value theorem and because both functions are differentiable exists $c$ such that: $$f'(c) = (f(x) - f(1))/(x - 1)$$ $$g'(c) = (g(x) - g(1))/(x - 1)$$ Then since $f(1) = g(1)$ I took everything to the side leaving only $f(1)$ and $g(1).$ I got $$f'(c)(x - 1)-f(x)= g'(c)(x - 1)-g(x)$$ so i have a relation between $f'(c)$ and $g'(c)$ but I can´t get to the desired result

Answer 1

This is FALSE.

Let $f(x)=x, g(x)=1$, we have

$$f(x)+x\le g(x)+x^2\Longleftrightarrow 0\le x^2-2x+1\Longleftrightarrow 0\le (x-1)^2$$

But

$$f'(1)+1=2\color{red}\neq0= g'(1)$$