Prove that $(s_2 - s_1) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) < s_n - s_1$

Question

Proposition:

Consider arbitrary positive real numbers $s_1,s_2,...s_n$ such that $s_1 < s_2 < s_3 \cdots < s_n$.

Prove that if $n$ is even and $n ≥ 4$, then $(s_2 - s_1) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) < s_n - s_1$

My attempt:

By induction:

Base case:

Consider arbitrary four numbers $s_1,s_2,s_3,s_4$ such that $s_1 < s_2 < s_3 < s_4$

Since $s_2 - s_3 < 0$, we have

$$\begin{align} s_4 - s_1 = s_4 - s_1 & \implies s_4 - s_1 + s_2 - s_3< s_4 - s_1 \\ & \implies (s_2 - s_1) + (s_4 - s_3) < s_4 - s_1 \end{align}$$

Induction step:

Suppose that for arbitrary even $n$, we have

$(s_2 - s_1) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) < s_n - s_1$

Now consider arbitrary two real values $s_{n+1},s_{n+2}$, such that

$$s_{n} < s_{n+1} < s_{n+2}$$

Then

$$\begin{align} & (s_1 - s_2) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) < s_n - s_1 \implies \\ & (s_1 - s_2) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) + {s_{n+2} - s_{n+1}}< s_{n+2} - s_{n+1} + s_n - s_1 < s_{n+2} - s_1 \end{align}$$

$\square$

Is it correct?

Answer 1

Your proof is true . But another way is: $s_n-s_1=\sum_{i=2}^{n}s_i-s_{i-1}>(s_2 - s_1) + (s_4 - s_3) + \cdots + (s_n - s_{n-1}) $

Answer 2

$$s_n-s_1=(s_n-s_{n-1})+(s_{n-1}-s_{n-2})+...(s_3-s_2)+(s_2-s_1)>$$ $$>(s_2-s_1)+(s_4-s_3)+...+(s_n-s_{n-1}).$$

Answer 3

I would reccomend a possibly more direct approach:

$$s_2-s_1 < s_3-s_1$$ $$s_4-s_3 < s_5-s_3$$ $$\ldots$$ $$s_{n-2}-s_{n-3} < s_{n-1}-s_{n-3}$$ $$s_n-s_{n-1}=s_n-s_{n-1}$$ And summing up all of these, you get

$$LHS < s_n-s_1$$