Let us divide our solution into $2$ cases, depending on if $f'(x)=0$ or $f'(x)\ne0$ for some $x\in\mathbb{R}$: Let us take the case where $f'(x)\ne 0.$ We see that $\left(\frac{df(x)}{dx}\right)^2+\frac{d^2f(x)}{dx^2}=0.$ Let $g(x)=\frac{df(x)}{dx}.$ We see that $\frac{dg(x)}{dx}=-g(x)^2.$ We can divide both sides by $g(x)^2$ to see $\frac{\frac{dg(x)}{dx}}{g(x)^2}=-1.$ Let us integrate both sides. We see that $\int \frac{\frac{dg(x)}{dx}}{g(x)^2}~dx=\int -1~dx=-x+C_0.$ Let us solve for $\int \frac{\frac{dg(x)}{dx}}{g(x)^2}~dx.$ Let $u=g(x).$ We see that $\int \frac{\frac{dg(x)}{dx}}{g(x)^2}~dx=\int \frac{1}{u^2}~du$ because $du=\frac{dg(x)}{dx}.$ We can see that $\int \frac{1}{u^2}~du=-\frac{1}{u}.$ Thus, $\int \frac{\frac{dg(x)}{dx}}{g(x)^2}~dx=-\frac{1}{g(x)}+C_1.$ Going back to the equation above, $-\frac{1}{g(x)}+C_1=-x+C_0.$ Let $C_0-C_1=-C.$ Thus $-\frac{1}{g(x)}=-x-C.$ We achieve the equation $g(x)=\frac{1}{x+C}.$ We can see that $\frac{df(x)}{dx}=g(x)=\frac{1}{x+C}.$ After integrating both sides of $\frac{df(x)}{dx}=\frac{1}{x+C},$ we achieve $f(x)+C_2=\int \frac{1}{x+C} dx.$ We know that $\int \frac{1}{x+C} dx=\ln(x+C)+C_3.$ Let $C_3-C_2=C_1.$ We see that $f(x)=\ln(x+C)+C_1.$ Let us take the case $f'(x)=0.$ We can see that $f(x)=C$ for any $C\in \mathbb{R}.$ Thus, $f(x)=\ln(x+C)+C_1$ when $f'(x)\ne0$ and $f(x)=C$ when $f'(x)=0.$

Atoms789 wrote: Let us divide our solution into $2$ cases, depending on if $f'(x)=0$ or $f'(x)\ne0$: ... We see that $\left(\frac{df(x)}{dx}\right)^2+\frac{d^2f(x)}{dx^2}=0.$ ... Thus, $f(x)=\ln(x+C)+C_1$ when $f'(x)\ne0$ and $f(x)=C$ when $f'(x)=0.$ I don't understand this post at all. - What does $f'(x)=0$ or $f'(x)\ne 0$ mean? Does that mean for some $x \in \mathbb{R}$, or for all $x$? Maybe $f'(x)=0$ for some $x$ and $f'(x) \not= 0$ for some other $x$ ? - We are only given the inequality $(f')^2 + f'' \le 0$, not the exact equality. - $\ln ( x + C)$ is not a solution, since it is not defined on all of $\mathbb{R}$. As for the solution: Let's write $f'=g$, then $g^2 + g' \le 0$. In particular, $g' \le 0$, so $g$ is decreasing. Assume first that there exists $a \in \mathbb{R}$ with $g(a) < 0$, then $g(x) < 0$ for $x \ge a$. For those $x$ we thus have $$ (1 / g ) ' ( x ) = - \frac{g'(x)}{g(x) ^2 } \ge 1.$$This would imply $\lim_{x \to \infty } 1 / g ( x ) = \infty$, in particular $1/g$ would be positive for large enough $x$, a contradiction. Next assume that there exists $a \in \mathbb{R}$ with $g(a) > 0$. Since $g$ is decreasing, we have $g(x) \ge g(a) > 0$ for $x \in (- \infty , a)$. Again we have $(1/g)' \ge 1$ for all $x \in ( - \infty , a )$ and thus $\lim_{x \to - \infty } 1 / g ( x ) = - \infty$. Then again $g$ would be negative for small enough $x$, a contradiction. So the only solution is $g \equiv 0$. Then $f$ is constant, which is indeed a solution.

@Gryphos, I must have accidentally solved for equality (my bad). Your solution seems to rule out the possibility of $f(x)$ being $\text{ln} (\left|x\right|+1)$ (which satisfies the inequality). Correct me if I am wrong.

Atoms789 wrote: @Gryphos, I must have accidentally solved for equality (my bad). Your solution seems to rule out the possibility of $f(x)$ being $\text{ln} (\left|x\right|+1)$ (which satisfies the inequality). Correct me if I am wrong. This function is not differentiable at $x=0$, so it does not satisfy the conditions.

Actually the solution is pretty straight-forward. One can consider the function \[g(x)=e^{f(x)}\]and see the that given condition can be written as \[g''(x)\le 0\]for all real numbers $x$. This means that $g$ is concave and lower-bounded by $0$, which can only be true if $g$ is constant (one can use that $g$ is concave and differentiable if and only if $g(x)\le g(y)+(x-y)g'(y)$ for all $x,y \in \mathbb{R}$). Finally, it follows that $f$ is also constant, and hence this is the only solution to the problem.

$(f^{\prime}(x))^2+f^{\prime\prime}(x)\leq 0$ $\implies e^{f(x)}(f^{\prime}(x))^2+e^{f(x)}f^{\prime\prime}(x)\leq0$ (since $e^{f(x)}>0$) $\implies \left(e^{f(x)}f^{\prime}(x)\right)^{\prime}\leq0$ $\implies \left(e^{f(x)}\right)^{\prime\prime} \leq0$ Setting $g(x)=e^{f(x)}$ gives $g^{\prime\prime}(x)\leq0$ $(*)$ and $g(x)>0$ $(**)$. Condition $(*)$ tells $g$ is convex. Claim: $g$ is constant. Proof: Suppose FTSOC, $g$ is not constant. So, $\exists a\in\mathbb{R}$ such that $g^{\prime}(a)\neq 0$. Also, $g(x)\leq g(a)+(x-a)g^{\prime}(a)$ (***) for all $x\in\mathbb{R}$ Case 1: If $g^{\prime}(a)>0$: Taking $x\rightarrow -\infty$ in (***) gives $g(x)$ has no lower bound (contradicting (**)) Case 2: If $g^{\prime}(a)<0$: Taking $x\rightarrow +\infty$ in (***) gives $g(x)$ has no lower bound (contradicting (**)) Hence $g$ must be constant. It follows from here that $f$ is also constant.