How to complete this derivative proof?

[Amad27]

Homework Statement

Assume that $f(x)$ has two derivatives in $(0,2)$ and $0<a<b<a+b<2$.

Prove that if $f(a)\ge f(a+b)$ and $f″(x)\le 0$ $\forall x \in (0, 2)$, then:

$$\frac{af(a)+bf(b)}{a+b} \ge f(a+b) \tag 1$$

Homework Equations

Below

The Attempt at a Solution

**MY PROOF:**

If $(1)$ is true then : $af(a) + bf(b) \ge af(a + b) + bf(a + b)$ , since it is given that: $af(a) \ge af(a+b)$ the objective is to prove: $f(b) \ge f(a + b)$

Assume $f(b) < f(a + b) $

$$\frac{f(a + b) - f(b)}{a} > 0 \tag2$$

For some $x$ the **Mean Value Theorem** guarantees for $\exists x_1 \in (b, a+b)$ that $$f'(x_1) = \frac{f(a + b) - f(b)}{a} $$

Hence, $(2)$ really means: $f'(x_1) > 0$ for some $x_1 \in (b, a+b)$

Since, $f(a) \ge f(a + b)$:

$$\frac{f(a + b) - f(a)}{b} \le 0 \tag3$$

By the **Mean Value Theorem,** $\exists x_2 \in (a, a+b)$ such that $$f'(x_2) =\frac{f(a + b) - f(a)}{b}$$

Hence $(3)$ really means, $$f'(x_2) \le 0$$

1. Obviously, if $f'(x_2) = 0$ then $f''(x_2)$ is a max, which means $f''(x_2) = 0$
2. Since $x_1 > x_2$ there $\exists x_3 \in (x_1, x_2)$ such that $f'(x_3) = 0$, which means $f''(x_3)$ is a max, meaning $f''(x_3) = 0$ contadictions.I just need help with 1 above, what if $x_2$ was a minimum?

Thanks!

 

[RUber]

If x2 is a minimum then f is constant if x2 is neither 0 or 2.

 

[SammyS]

Homework Statement

Assume that $f(x)$ has two derivatives in $(0,2)$ and $0<a<b<a+b<2$.

Prove that if $f(a)\ge f(a+b)$ and $f″(x)\le 0$ $\forall x \in (0, 2)$, then:
$$\frac{af(a)+bf(b)}{a+b} \ge f(a+b) \tag 1$$

Homework Equations

Below

The Attempt at a Solution

**MY PROOF:**

If $(1)$ is true then : $af(a) + bf(b) \ge af(a + b) + bf(a + b)$ , since it is given that: $af(a) \ge af(a+b)$ the objective is to prove: $f(b) \ge f(a + b)$

Assume $f(b) < f(a + b)$
$$\frac{f(a + b) - f(b)}{a} > 0 \tag2$$
For some $x$ the **Mean Value Theorem** guarantees for $\exists x_1 \in (b, a+b)$ that $$f'(x_1) = \frac{f(a + b) - f(b)}{a} $$
Hence, $(2)$ really means: $f'(x_1) > 0$ for some $x_1 \in (b, a+b)$

Since, $f(a) \ge f(a + b)$:
$$\frac{f(a + b) - f(a)}{b} \le 0 \tag3$$
By the **Mean Value Theorem,** $\exists x_2 \in (a, a+b)$ such that $$f'(x_2) =\frac{f(a + b) - f(a)}{b}$$
Hence $(3)$ really means, $$f'(x_2) \le 0$$
1. Obviously, if $f'(x_2) = 0$ then $f''(x_2)$ is a max, which means $f''(x_2) = 0$
2. Since $x_1 > x_2$ there $\exists x_3 \in (x_1, x_2)$ such that $f'(x_3) = 0$, which means $f''(x_3)$ is a max, meaning $f''(x_3) = 0$ contadictions.I just need help with 1 above, what if $x_2$ was a minimum?

Thanks!

Use $\small{\text{$ ___ $}}$ for delimiters for inline LaTeX.

I fixed up some of it.

 

[Amad27]

If x2 is a minimum then f is constant if x2 is neither 0 or 2.

@RUber, is my proof correct though?