proving $1+ \bigg(\int|f|d\mu\bigg)^2 \le \bigg(\int \sqrt{1+|f|^2}d\mu\bigg)^2\le\bigg(1+\int|f|d\mu\bigg)^2$ Let $f \in L^1(X, \mu)$, with $\mu(X) = 1$.  Prove that
$$1+\bigg(\int|f|d\mu\bigg)^2\le\bigg(\int\sqrt{1+|f|^2}d\mu\bigg)^2\le\bigg(1+\int|f|d\mu\bigg)^2$$
$\textbf{My attempt}$:


*

*for the first inequality (I stuck at the end for the first inequality and not sure if this is going anywhere)
\begin{align}
1+\bigg(\int|f|d\mu\bigg)^2 
& \le \bigg(1+\int|f|d\mu\bigg)^2 = \bigg(\int(1+|f|)d\mu\bigg)^2\\
& =\bigg(\int\sqrt{(1+|f|)^2}d\mu\bigg)^2=\bigg(\int\sqrt{1+|f|^2+2|f|} d\mu\bigg)^2 \\
& \le (????)\\
\end{align}

*for the second inequality (is this correct?)
\begin{align}
\bigg(\int\sqrt{1+|f|^2}d\mu\bigg)^2 
& =\bigg(\int\sqrt{(1+|f|)^2-2|f|}d\mu\bigg)^2 \le \bigg(\int\sqrt{(1+|f|)^2}d\mu\bigg)^2\\
& = \bigg(\int (1+|f|)d\mu\bigg)^2\\
& \le \bigg(\mu(X) + \int|f|d\mu\bigg)^2 =\bigg(1+\int|f|d\mu\bigg)^2
\end{align}
 A: The second inequality is fine.
For the first inequality, first note that the desired inequality can be expressed as $\sqrt{1+(\mathbb{E}(|f|))^2}\le \mathbb{E}\sqrt{1+|f|^2}$, which then can be seen to be a simple consequence of the Jensen's inequality since the function $\sqrt{1+x^2}$ is convex.
A: The second one is correct. For the first one consider $a+b\int |f|d\mu =\int (a+b|f|)d\mu \leq \int \sqrt {a^{2}+b^{2}} \sqrt {1+|f|^{2}} d\mu$. If $a^{2}+b^{2} \leq 1$ this gives $a+b\int |f|d\mu  \leq \int \sqrt {1+|f|^{2}} d\mu$. Maximize LHS over all $a,b$ with $a^{2}+b^{2} \leq 1$  to get $\sqrt {1+(\int|f|d\mu)^{2}} \leq \int \sqrt {1+|f|^{2}} d\mu$.
Actually maximization occurs when $a =\frac 1 {\sqrt {1+\int (|f|d\mu)^{2}}}$ and $b =\frac {\int |f|d\mu} {\sqrt {1+\int (|f| d\mu)^{2}}}$
A: We'll need to use the fact that, by Cauchy-Schwartz:
$$1+\alpha\beta = 1\times 1 +\beta\times\gamma 
 \leq\sqrt{1+\beta^2}\sqrt{1+\gamma^2}\tag{1}$$
First inequality:
$$\begin{split}
1+\bigg(\int|f|d\mu\bigg)^2&=\int\bigg[1\bigg(\int|f(x)|d\mu(x)\bigg)|f(y)|\bigg]d\mu(y)\\
&\leq\int\sqrt{1+\bigg(\int|f(x)|d\mu(x)\bigg)^2}\sqrt{1+|f(y)|^2}d\mu(y) \,\,\,\,\,\mbox{  using }(1)\\
&=\sqrt{1+\bigg(\int|f|d\mu\bigg)^2}\cdot\int\sqrt{1+|f|^2}d\mu
\end{split}$$
Second inequality:
$$
\bigg(\int\sqrt{1+|f|^2}d\mu\bigg)^2  \leq\bigg(\int\sqrt{1+2|f| + |f|^2}d\mu\bigg)^2 = \bigg(\int(1+|f|)d\mu\bigg)^2= \bigg(1 + \int|f|d\mu\bigg)^2
$$
