$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.

The inequality in question is $$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$ for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.

Suppose that this is true. Take any $t\in(0,1)$. For all $x=(x_1,x_2)\in\R^2$, let $f(x):=f_t(x):=y+tx$, where $y:=(-x_2,x_1)$. Then $f$ is $2$-Lipshitz.

For each $x\in\R^2$, let now $u=f(x)$. Then \eqref{1} becomes $$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$ Letting here $|x|\to\infty$ and diving by $t$, we get $$2(1+t^2)\le\frac C\ep\,t$$ for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$

$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.

The inequality in question is $$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$ for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.

To obtain a contradiction, suppose that this is true. Take any $t\in(0,1)$. For all $x=(x_1,x_2)\in\R^2$, let $f(x):=f_t(x):=y+tx$, where $y:=(-x_2,x_1)$, so that $|y|=|x|$ and $y$ is orthogonal to $x$. Then $f$ is $2$-Lipshitz.

For each $x\in\R^2$, let now $u=f(x)$. Then \eqref{1} becomes $$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$ For any nonzero $x\in\R^2$, dividing both sides of this inequality by $t|x|^4$ and then letting $|x|\to\infty$, we get $$2(1+t^2)\le\frac C\ep\,t$$ for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$

$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.

The inequality in question is $$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$ for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.

Suppose that this is true. For any real $t>0$ and any $x=(x_1,x_2)\in\R^2$, let $u=f(x):=y+tx$, where $y:=(-x_2,x_1)$. Then \eqref{1} becomes $$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$ Letting here $|x|\to\infty$ and diving by $t$, we get $$2(1+t^2)\le\frac C\ep\,t$$ for all real $t$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$

$\newcommand\R{\mathbb R}\newcommand\ep{\varepsilon}$Such a statement does not hold for any $n\ge2$. Indeed, in view of the natural embedding of $\R^2$ into $\R^n$ for $n\ge2$, without loss of generality $n=2$.

The inequality in question is $$2u^\top f(x)u^\top x\le\ep|u|^2+\frac C\ep\,(u^\top x)^2 \tag{1}\label{1}$$ for all $x,u$ in $\R^2$, where $|\cdot|$ is the Euclidean norm.

Suppose that this is true. Take any $t\in(0,1)$. For all $x=(x_1,x_2)\in\R^2$, let $f(x):=f_t(x):=y+tx$, where $y:=(-x_2,x_1)$. Then $f$ is $2$-Lipshitz.

For each $x\in\R^2$, let now $u=f(x)$. Then \eqref{1} becomes $$2t(1+t^2)|x|^4\le\ep(1+t^2)|x|^2+\frac C\ep\,t^2 |x|^4.$$ Letting here $|x|\to\infty$ and diving by $t$, we get $$2(1+t^2)\le\frac C\ep\,t$$ for all $t\in(0,1)$. Letting now $t\downarrow0$, we get $2\le0$. $\quad\Box$