Infimum for the values $\int_{\mathbb R^n} u^2-\ln(1+u^2)\text dx$ Is there any way in the course of Real analysis (For example, Zygmund's Measure Theory, Stein's Real analysis) to obtain the following extreme values 
\begin{equation}
\text{inf}\left\{\int_{\mathbb R^5}u^2-\ln(1+u^2)\text dx\,\,\right|\,\left.\ \int_{\mathbb R^5} u^{14}=1\right\}
\end{equation} 
and those $u$ which take exactly the infimum if it exists?
 A: Let $f(x) = x - \ln(1+x)$, and let $I[u] = \int_{\mathbb{R}^5}{f(u^2(x))\,dx}$ (defined, say, for any $u\in L^{14}$ for which this integral is finite). Notice that $f$ is increasing on $x\ge 0$, and $f(x)\ge 0$ for $x\ge 0$, with equality iff $x=0$. It follows that $I[u]\ge 0$ for all $u$, and furthermore $I[u]>0$ if $\int{|u|^{14}} = 1$, since $I[u] = 0 \implies f(u^2) = 0$ a.e. $\implies u = 0$ a.e..
So we know that $\inf \{I[u]\,|\int{|u|^{14}} = 1\}\ge 0$. A natural question to ask is whether in fact $\inf \{I[u]\,|\int{|u|^{14}} = 1\} = 0$. I argue that the $\inf$ is indeed $0$, and hence is not achieved. To show this, we use a scaling argument. Let $u$ be sufficiently nice (say smooth and compactly supported), so that $u\in L^2(\mathbb{R}^5)\cap L^{14}(\mathbb{R}^5)$, and normalize $u$ so that $\int{|u|^{14}} = 1$. For $\epsilon>0$, let
$$u_{\epsilon}(x) = \epsilon^{-5/14}u\left(\frac{x}{\epsilon}\right).$$
Defining the $u_{\epsilon}$ in this way gives us that
$$\int_{\mathbb{R}^5}{|u_{\epsilon}|^{14}} = \int_{\mathbb{R}^5}{|u|^{14}} = 1. $$ 
Now, notice that
$$ I[u_{\epsilon}] = \int_{\mathbb{R}^5}{[u_{\epsilon}^2 - \ln(1+u_{\epsilon}^2)]}\le\int_{\mathbb{R}^5}{u_{\epsilon}^2} $$
and
$$ \int_{\mathbb{R}^5}{u_{\epsilon}^2(x)\,dx} = \int_{\mathbb{R}^5}{\epsilon^{-5/7}u^2\left(\frac{x}{\epsilon}\right)\,dx} = \int_{\mathbb{R}^5}{\epsilon^{30/7}u^2(x)\,dx}\xrightarrow{\epsilon\rightarrow 0} 0 .$$
Taking a sequence $\epsilon_k\rightarrow 0$ gives that $\{u_{\epsilon_k}\}$ is a sequence of functions with $\int{|u_{\epsilon_k}|^{14}} = 1$ and $I[u_{\epsilon_k}]\rightarrow 0$, thus showing that $\inf \{I[u]\,|\int{|u|^{14}} = 1\} = 0$, as desired.
