How to prove this inequality that i found while proving convergence of steepest descent for quadratic case

Question

Let $0<y_1\le y_2 \ldots \le y_n$.

Also, $\sum_{i=1}^n x_iy_i=\alpha y_1+(1-\alpha)y_n$ where $\alpha\ge 0$, $x_i \ge 0$ and $\sum x_i=1$.

To show $\sum_{i=1}^n x_i/y_i \le \alpha/y_1 +(1-\alpha)/y_n$

Answer 1

(Fill in the gaps as needed. If you're stuck, show your work)

Define $\alpha_i$ such that $y_i = \alpha_i y_1 + (1-\alpha_i) y_n$. Note that $ 0 \leq \alpha_i \leq 1$.

1. Show that $$ \alpha = \sum x_i \alpha_i.$$

Substitute in and show that it holds.

2. Show that $$ \frac{1}{ y_i} = \frac{1}{ \alpha_i y_1 + (1-\alpha_i) y_n} \leq \frac{ \alpha_i} { y_1 } + \frac{ (1-\alpha_i) } { y_n}.$$

Apart from expanding everything, this follows from $ f(x) = \frac{1}{x} $ is a convex function on $x> 0$.

3. Show that $$\sum \frac{x_i}{ y_i} \leq \frac{ \alpha } { y_1 } + \frac{ (1-\alpha) } { y_n}.$$

Substitute in and show that it holds.