Homework on basic inequalities. Let $a_j$ be a sequence of positive reals. Show that
(a) $\left(\sum_{j=1}^\infty a_j\right)^\theta \le \sum_{j=1}^\infty a_j^\theta$ for any $0\le\theta\le1$.
(b) $\sum_{j=1}^\infty a_j^\theta \le \left(\sum_{j=1}^\infty a_j\right)^\theta $ for any $1\le\theta<\infty$.
(c) $\left(\sum_{j=1}^N a_j\right)^\theta \le N^{\theta-1} \sum_{j=1}^N a_j^\theta$, when $1\le\theta<\infty$.
(d) $\sum_{j=1}^N a_j^\theta \le \sum_{j=1}^N N^{1-\theta} \left(\sum_{j=1}^N\right)^\theta$, when $0\le\theta\le1$.
I think it is about the convex/concave function but I don't have an explicit idea.
 A: For (a) note that $(x+y)^{p}\le x^{p}+y^{p}$ when $0<p<1$ and $x,y\ge0$. Then:
$(\sum_{k=1}^{n}a_{k})^{\theta}\le\sum_{k=1}^{n}a_{k}^{\theta}\le\sum_{k=1}^{\infty}a_{k}^{\theta}$
where $n$ was arbitrary. To prove the first inequality mentioned notice that for $y=0$ it is clear and it suffices to show the case when $y=1$ since if it is true in this case then: $$(x+y)^{p}=y^{p}(\frac{x}{y}+1)^{p}\le y^{p}((\frac{x}{y})^{p}+1)=x^{p}+y^{p}$$ Then simply show that the function $f(x)=(x+1)^{p}-x^{p}$ is decreasing on the non-negative reals. For $(b)$ note that $x^{p}+y^{p}\le(x+y)^{p}$ when $p\ge1$ and $x,y\ge0$ and we proceed by a similar proof as in (a). (c) follows from Holder's inequality:
$\sum_{k=1}^{N}a_{k}\le(\sum_{k=1}^{N}1)^{\frac{\theta-1}{\theta}}(\sum_{k=1}^{N}a_{k}^{\theta})^{\frac{1}{\theta}}=(N^{\theta-1}\sum_{k=1}^{N}a_{k}^{\theta})^{\frac{1}{\theta}}$
and (d) follows from the reverse Holder's inequality:
$\sum_{k=1}^{N}a_{k}\ge(\sum_{k=1}^{N}1)^{\frac{1}{\theta}(\theta-1)}(\sum_{k=1}^{N}a_{k}^{\theta})^{\frac{1}{\theta}}=(N^{\theta-1}\sum_{k=1}^{N}a_{k}^{\theta})^{\frac{1}{\theta}}$
