I am looking for a proof, reference, comment of an inequality as follows:

If $f(x)$ and $g(x)$ be two continuous derivative funcions in interval $[a, b]$. Such that:

1. $f(a)=g(a)$ and $f(b)=g(b)$

2. $($$f''(x) > 0$ or $f''(x) < 0$$)$ and $($$g''(x) > 0$ or $g''(x) < 0$$)$

3. $$|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)+f(a)|\ge|g(x)-\frac{g(b)-g(a)}{b-a}(x-a)+g(a)|$$

Then

$$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$

I am looking for a proof, reference, comment of an inequality as follows:

If $f(x)$ and $g(x)$ be two continuous derivative funcions in interval $[a, b]$. Such that:

1. $f(a)=g(a)$ and $f(b)=g(b)$

2. $($$f''(x) > 0$ or $f''(x) < 0$$)$ and $($$g''(x) > 0$ or $g''(x) < 0$$)$

3. $$|f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a)|\ge|g(x)-\frac{g(b)-g(a)}{b-a}(x-a)-g(a)|$$

Then

$$\int_{a}^{b}{ \sqrt{1+f'(x)^2}}dx \ge \int_{a}^{b}{ \sqrt{1+g'(x)^2}}dx$$