I've started to read Rao's Theory of Orlicz Spaces book. There are two points I could not get well at proof of Vallée Poussin's uniform integrablity theorem. Please find the theorem above.
The two points I referred are in $(ii) \Rightarrow (iii)$. I know they are so easy but I need to see them clearly for understanding. 
First, I cannot see why $$\varphi(k)\frac{x-k}{x}\leq \frac{\Phi(x)}{x} \ \ \ \text{when} \ \ k<x \ \ \text{and} \ \ k\to \infty$$
Second, I couldn't show that $$\Phi(\alpha x+\beta y) = \int_0^{|\alpha x + \beta y|} \varphi(t)dt\leq \alpha\int_0^{|x|} \varphi(t)dt + \beta\int_0^{|y|} \varphi(t)dt = \alpha\Phi(x)+\beta\Phi(y)$$
Thanks in advance for any explanation.
For the first part, this is two separate statements: first that for $k<x$, $\phi(k)\frac{x-k}{x}\leq\frac{\Phi(x)}{x}$, and second, that as $k\to\infty$ the LHS $\to\infty$ so that the RHS must as well.
For the second statement, we have already that $\phi(n)\to\infty$ with $n$. Then for any $k$, $$\lim_{k\to\infty}\lim_{x\to\infty} \phi(k)\frac{x-k}{x} = \lim_{k\to\infty} \phi(k)=1.$$
The first claim is a little weirder. But note that $\Phi(x)/x$ is exactly the mean value of $\phi(k)$ on the interval $[0,x]$, and also that $\frac{x-k}{x}=\frac{\lvert[k,x]\rvert}{\lvert[0,x]\rvert}$ is the fraction of the interval remaining once you've reached $k$. Recall that $\phi$ is increasing, and think geometrically: $\phi(k)\frac{x-k}{x}$ must be an underestimate of the mean on the whole interval, since $$\phi(k)\frac{x-k}{x}=\frac{\phi(k)}{x}\int_k^x 1\,dt\leq\frac{1}{x}\int_k^x \phi(t)\,dt\leq \frac{1}{x}\int_{0}^x\phi(t)\,dt=\frac{\Phi(x)}{x}.$$
Here, in the first inequality we have used that $\phi$ is increasing, and in the second that it is positive.
OK, now let's deal with the convexity. Note that since $$\Phi'(x)= \operatorname{sgn}(x)\,\phi(\lvert x \rvert),$$ then for $x\neq0$ the second derivative $$\Phi'(x)=0+(\operatorname{sgn}(x))^2\,\phi'(\lvert x \rvert)$$ is always positive, since $\phi'$ is positive.