Help Solving an Inequality Can you help me solve the following problem of inequality? The context from where this arose, is I think unnecessary to be discussed here, and it is just a step which is preventing me from solving a problem. So, here it is:
Suppose that $a_i$, $b_i$ and $d_i$ $(1\leq i\leq n)$ are real numbers, with the $d_i$'s being positive, satisfying: $$\sum_{i=1}^n a_i^2 \leq \alpha \hspace{1cm}\textrm{and}\hspace{1cm}\sum_{i=1}^n \frac{b_i^2}{d_i} \leq \alpha~,$$  for some non-negative $\alpha$. Then, is the following true for any $\theta \in [0,1]$?
$$\sum_{i=1}^n \frac{(\theta a_i + (1-\theta)b_i)^2}{\theta + (1-\theta)d_i} \leq \alpha~?$$ 
My attempts were to show that the function $$f(\theta) := \sum_{i=1}^n \frac{(\theta a_i + (1-\theta)b_i)^2}{\theta + (1-\theta)d_i}$$ is convex, or at least quasiconvex, which would solve the problem, since $f(1)= \sum_{i=1}^n a_i^2 $ and $f(0) = \sum_{i=1}^n \frac{b_i^2}{d_i}$. Unfortunately, I can show neither.
Any help will be greatly appreciated!!

By CS inequality,
$$\alpha =\theta \alpha +(1-\theta)\alpha\geqslant \theta \sum a_i^2 + (1-\theta)\sum \frac{b_i^2}{d_i} = \sum \left(\frac{\theta^2 a_i^2}{\theta} + \frac{(1-\theta)^2b_i^2}{(1-\theta)d_i}\right)\\ \geqslant \sum\frac{(\theta a_i + (1-\theta)b_i)^2}{\theta+(1-\theta)d_i}$$
