A rough estimate for $ F(\mu):=\Bigg[\frac{4}{\pi}+\frac{\sqrt{2}}{\pi}\bigg(8+\pi\log K+\frac{\pi}{2}\log(1+\mu)\bigg)^{1/2}\Bigg](1+\mu)^{1/2}K$ 
Suppose $\mu\geq 0$ and define
  $$
F(\mu):=\Bigg[\frac{4}{\pi}+\frac{\sqrt{2}}{\pi}\bigg(8+\pi\log K+\frac{\pi}{2}\log(1+\mu)\bigg)^{1/2}\Bigg](1+\mu)^{1/2}K
$$
  where $K>1$ is some constant. Then there exists $c'>0$ and $c>0$ such that
  $$
F(\mu)\leq cK^2\log K
$$
  for all $\mu\leq c'K^2\log(c'K)$.

The argument above is made without proof for estimating an integral in a paper I read. Could anyone elaborate how one should choose the constants $c$ and $c'$?

I have tried to use Cauchy-Schwarz to isolate the factor $K$ and the other terms, but cannot match the exponents in $K^2\log K$. 
 A: WLOG we can assume that $\mu>1$ and $\log\mu>1$.
Then
\begin{align}
F(\mu)=&c_1(1+\mu)^{1/2}K
+\big[c_2+c_3\log K+c_4\log(1+\mu)\big]^{1/2}(1+\mu)^{1/2}K\\
\leq &c_1(2\mu)^{1/2}K
+[c_2+c_3\log K]^{1/2}(2\mu)^{1/2}K+[c_4\log(2\mu)]^{1/2}(2\mu)^{1/2}K\\
\leq &c_5\mu^{1/2}K+c_6(\mu\log K)^{1/2}K+c_7(\mu\log\mu)^{1/2}K\\
\leq &c_8(\mu\log\mu)^{1/2}K+c_6(\mu\log K)^{1/2}K+c_7(\mu\log\mu)^{1/2}K
\end{align}
Suppose $\mu\geq cK^2\log(cK)$.
All we need to do now is to estimate the following two terms 


*

*$(\mu\log K)^{1/2}K$

*$(\mu\log\mu)^{1/2}K$


For the first one
$$
(\mu\log K)^{1/2}K
\leq (\log K)^{1/2}[cK^2\log(cK)]^{1/2}K\leq c_9K^2\log K.
$$
For the second one, denote $G(\mu):=(\mu\log\mu)^{1/2}K$. Then
$$
G(\mu)\leq
\biggr[cK^2\log(cK)\cdot \log\big(cK^2\log(cK)\big)\biggr]^{1/2}K
\leq \sqrt{c}K^2\biggr[\log(cK)\cdot \log\big(cK^2\log(cK)\big)\biggr]^{1/2}\tag{1}
$$
Note that
$$
\log\big(cK^2\log(cK)\big)=\log(cK^2)+\log\big(\log(cK)\big)
\leq2\log(\sqrt{c}K)+\log(cK)\leq c_{10}\log(c_{10}K).
$$
Thus 
$$
\log(cK)\cdot \log\big(cK^2\log(cK)\big)\leq c_{11}^2[\log(c_{11}K)]^2\tag{2}
$$
Combining (1) and (2), we have
$$
G(\mu)\leq c_{12}K^2\log(c_{12}K)
=c_{12}K^2(\log c_{12}+\log K)\leq c_{13}K^2\log K.
$$
