Inequality for Euclidean norm Let:| | be  Euclidean norm on $\mathbb{R}^{n}$ and 
 $b :  \mathbb{R}^{n}\longmapsto \mathbb{R}^{n}$ and $\sigma :  \mathbb{R}^{n}\longmapsto \mathbb{R}^{n\times m}$  two continuous functions.  
Suppose there exist a constant $C_{1}$ such that :
$$ |b(x)|+|\sigma(x)| \leq C_{1}(1+|x|) \forall  x \in \mathbb{R}^{n}. $$
How we can show that this implies that , for fixed $p \geq $2, we can find a constant $C_{2} > 0$ such that :
$$
 p |x|^{p-2}x.b(x)+\frac{1}{2}p(p-1)|x|^{p-2}|\sigma(x)|^{2}\leq C_{2}(1+|x|^{p}) \forall x \in \mathbb{R}^{n}?
$$
you can find my question here reference
I'll be grateful for any help offered!
 A: For $p\ge2$ and $u\ge0$, Jensen's inequality gives
$$
\begin{align}
\left(\frac{1+u^2}{2}\right)^{p/2}&\le\frac{1+u^p}{2}\tag{1}\\
\left(1+u^2\right)^{p/2}&\le2^{p/2-1}\left(1+u^p\right)\tag{2}
\end{align}
$$
Thus,
$$
\begin{align}
\frac{u^{p-2}(1+u^2)}{1+u^p}
&=\left(\frac{u^p}{1+u^p}\right)^{1-2/p}
\left(\frac{(1+u^2)^{p/2}}{1+u^p}\right)^{2/p}\tag{3}\\
&\le1^{1-2/p}\left(2^{p/2-1}\right)^{2/p}\tag{4}\\
&=2^{1-2/p}\tag{5}
\end{align}
$$
Explanation:
$(4)$: $\frac{u^p}{1+u^p}\le1$ and $(2)$
Then
$$
\begin{align}
&p|x|^{p-2}x\cdot b(x)+\tfrac12p(p-1)|x|^{p-2}|\sigma(x)|^2\tag{6}\\
&=p|x|^{p-2}\left(x\cdot b(x)+\tfrac12(p-1)|\sigma(x)|^2\right)\tag{7}\\
&\le p|x|^{p-2}\left(|x|\,|b(x)|+\tfrac12(p-1)C_1^2(1+|x|)^2\right)\tag{8}\\
&\le p|x|^{p-2}\left(C_1(1+|x|)^2+\tfrac12(p-1)C_1^2(1+|x|)^2\right)\tag{9}\\
&\le p|x|^{p-2}\left(C_1+\tfrac12(p-1)C_1^2\right)(1+|x|)^2\tag{10}\\
&\le p(C_1+\tfrac12(p-1)C_1^2)2^{1-2/p}(1+|x|^p)\tag{11}
\end{align}
$$
Explanation:
$\ \ (7)$: factor out $|x|^{p-2}$
$\ \ (8)$: apply Hölder to $x\cdot b(x)$ and $|\sigma(x)|\le C_1(1+|x|)$
$\ \ (9)$: use $|x|\le1+|x|$ and $|b(x)|\le C_1(1+|x|)$
$(10)$: factor out $(1+|x|)^2$
$(11)$: $(5)$
So set $C_2=p(C_1+\tfrac12(p-1)C_1^2)2^{1-2/p}$
