I claim $f(x)=0$ for all $x$ is the only solution. Let $P(x,y)$ denotes the given assertion, assume the contrary that there is an $f\not\equiv 0$. I first show there is a $y_0$ such that $f(y_0)>0$. Suppose it is not the case, take any $y_0$ with $f(y_0)<0$. $P(-\epsilon,y_0-\epsilon^2)$ then gives \[ 0>f(y_0) \ge \left(1-\frac{1}{\epsilon}\right)f\bigl(y_0-\epsilon^2\bigr). \]In particular if $\epsilon>1$, then $f(y_0-\epsilon^2)>0$. Having established this, we now show $f$ must identically be zero. Fix $\epsilon>0$. Notice that \[ f\left(y_0+\epsilon^2\right)\ge \left(1+\frac{1}{\epsilon}\right)f(y_0). \]In particular, $f(y_0+\epsilon^2)>0$. Iterating, we find that for every $n\in\mathbb{N}$, \[ f\left(y_0+n\epsilon^2\right)\ge \left(1 + \frac{1}{\epsilon}\right)^n f(y_0). \]We are now in business. Choose $\epsilon = n^{-1/2}$. We then get \[ f(y_0 + 1) \ge \bigl(\sqrt{n}+1\bigr)^n f(y_0). \]Since the RHS is unbounded, we reached a contradiction.

Setting $x=-1$ gives that $f(z) \geq 0$ for all $z\in \mathbb{R}$ and hence $f(x^2 + y) \geq f(y)$ for all $x>0$, i.e. $f$ is increasing. Let $k\in \mathbb{R}$ be arbitrary. For $y<k$ and any $x\in (0,\sqrt{k-y}]$ we have $f(x^2 + y) \leq f(k)$, but also $f(x^2 + y) \geq (\frac{1}{x}+1)f(y) \geq \frac{f(y)}{x}$. If we suppose, for contradiction, that $f(y) > 0$ for some $y<k$, then $f(k) > 0$ and $x=\frac{f(y)}{2f(k)}$ yield $f(k) \geq f(x^2+y) \geq \frac{f(y)}{x} = 2f(k)$, contradicting $f(k) > 0$. Therefore $f(y) = 0$ for all $y<k$. Since $k$ was arbitrary, this shows that $f\equiv 0$ (and this function satisfies the given condition).

Setting $x=-1$ we get: $$f{\left(1+y\right)} \geq 0 \Longrightarrow \forall_{y \in \mathbb{R}}\; f(y) \geq 0$$Next set $x=\frac{1}{\sqrt{n}}$ for $n \in \mathbb{N}$ then we get: $$f{\left(y+\frac{1}{n}\right)} \geq \left(1+\sqrt{n}\right) f(y) \Longrightarrow f(y+1) \geq \left(1+\sqrt{n}\right)^n f(y)$$where the second result comes from applying the first inequality $n$ times. If $f(y)>0$ then as $n \rightarrow \infty$, $\mathrm{RHS} \rightarrow \infty$ which cannot happen as $f(y+1)$ is fixed. Thus in fact $f \equiv 0$ which is indeed a solution.

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x^2 + y) \ge \left(\frac{1}{x} + 1\right)f(y)$$holds for all $x \in \mathbb{R} \setminus \{0\}$ and all $y \in \mathbb{R}$. We claim the only solution is $\boxed{f\equiv 0}$. This clearly works, we now prove it is the only solution. Claim: $f(x)\ge 0 $ for each $x\in \mathbb{R}$. Proof: $P(-1,x-1): f(x)\ge 0$. $\blacksquare$ $P\left(\frac{1}{\sqrt{x}},y\right): f\left(\frac{1}{x}+y\right)\ge (\sqrt{x}+1)f(y)$. Claim: For any positive integer $k$, $f\left(\frac{k}{x}+y\right)\ge (\sqrt{x}+1)^k f(y)$ for all reals $x,y$. Proof: Induct on $k$. Base case $k=1$ works. Suppose $f\left(\frac{k-1}{x}+y\right)\ge (\sqrt{x}+1)^{k-1}f(y)$ for all reals $x,y$. Then \begin{align*} f\left(\frac{k}{x}+y\right) \\ =f\left(\frac{1}{x}+\left(\frac{k-1}{x}+y\right)\right) \\ \ge (\sqrt{x}+1)f\left(\frac{k-1}{x}+y\right) \\ \ge (\sqrt{x}+1)^k f(y), \\ \end{align*}as desired. $\blacksquare$ So for each positive integer $n$ and real number $y$, $f(y+1)\ge (\sqrt{n}+1)^n f(y)$. If $f$ is not identically zero, set $y$ with $f(y)>0$ and $n$ sufficiently large. Since $f(y+1)$ is fixed, it follows that \[(\sqrt{n}+1)^n f(y)>f(y+1),\]a contradiction.

Case $1 : f$ is constant. Let $f(x) = c$ then we have $c \ge \frac{xc+c}{x} \implies c = 0$ so $f(x) = 0$. Case $2 : f$ isn't constant. $P(-1,y) : f(y+1) \ge 0 \implies f(x) \ge 0$ for $x \in \mathbb{R}$. Note that applying $P(\sqrt{\frac{1}{k}},y)$ for $k$ times gives that $\frac{f(y+1)}{f(y)} \ge (1+\frac{k}{\sqrt{k}})^k$ which isn't true for some large $k$ cause $\frac{f(y+1)}{f(y)}$ is fixed for a fixed $y$ but $(1+\frac{k}{\sqrt{k}})^k$ isn't so for some large $k$ RHS will be larger so there exists no such $f$. Answers $: f(x) = 0$

Let $P(x,y)$ be the assertion. Firstly, $P(-1, y)$ gives that $f(y+1) \geq 0$ implying that the range of $f$ is non-negative reals only. Assume $f(y) \neq 0$ for some $y$. We further have, $P(\epsilon, y)$ where $\epsilon \to 0$ giving $f(y+\epsilon^2) \geq \left( \frac{1}{\epsilon}+1 \right) f(y)$. Since the right side goes to infinity, we have $f(y+\delta) \to \infty$ for $ \delta \to 0$. Therefore, there exists $\delta$ for which $f(y+ \delta) > f(y+1)$. Finally, $$P(\sqrt{1- \delta}, y+ \delta) \implies f(y+1) \geq \left( \frac{1}{\sqrt{1- \delta}}+1 \right)f(y+ \delta)$$, a contradiction.

We claim that $f\equiv 0.$ It clearly works. Let $P(x,y)$ denote the given assertion. $P(1,x-1)$ implies $f(x)\geq 0.$ $P(x^{-1},y)$ yields $f(y+\tfrac{1}{x^2})\geq f(y)(x+1)$ so by induction $f(y+1)\geq f(y)(x+1)^{x^2}.$ We can $y$ arbitrarily large so that $f(y)\leq 0,$ as claimed.

Uhm, am I doing something wrong? If I am not having a strok,letting $x=\sqrt{\frac{1}{k}}$ is unnecessary. Let $P(x,y)$ be the assertion of the given equations. By $P(-1,y)$, we can conclude the range is $\mathbb{R}_{\ge 0}$. Let $x \rightarrow 0$. We can conclude $f(y) \ge f(y) \cdot big \ stuff$. Which gives that either $f(x)$ is always negative, or $f(x)=0$. But since range is $\ge0$, we can conclude $f \equiv 0 \ \blacksquare$ Note: Solution is wrong since $f$ is not proven continous

Letting $x=-1$, we get $f(y+1)\ge 0$ for all $y$. Now, let $f(y+a)\ge \left(\frac{1}{a}+1\right)f(y)$, and so by induction, $f(y+ab)\ge \left(\frac{1}{a}+1\right)^bf(y).$ Now, \[f(y+1)=f\left(y+n\cdot \frac{1}{n}\right)\ge (n+1)^nf(y)\]for positive integers $n$. If $f(y)\neq 0$ then making $n$ sufficiently large makes that false. Therefore, $f(y)=0$ which works. edit: frick missed an exponent. well. /shrug. proof should be similar enough anyway.

$f(x)=0$ is an only solution. It is easy to see that $f(x)=0$ satisfy the condition. Setting $x=-1$ gives that $f(y)\ge 0$ and $f$ is increasing. Let $x\leq 1$ and we get that $$f(y+1)\ge f(x^2+y)\ge (\frac{1}{x} + 1)f(y)$$If $f(y+1)\neq 0$ then making $x$ very small makes that false, so $f(y)=0$ for all $y \in \mathbb{R}$.

Lukaluce wrote: Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(x^2 + y) \ge (\frac{1}{x} + 1)f(y)$$holds for all $x \in \mathbb{R} \setminus \{0\}$ and all $y \in \mathbb{R}$. for $x=-1$ we get $f(y+1)\geqslant 0$ Using indaction we get: $f(y+nx^2)\geqslant (\frac{1}{x}+1)^nf(y)$ For $x=\frac{1}{n}$ and using idaction we get: $f(y+\frac{m}{n})\geqslant (\frac{1}{n}+1)^{mn}f(y)$ Taking $m=n$ and $n\rightarrow +\infty $ we get $f(x)=0$ is the only sollution

We claim the only solution is $f(x) \equiv 0$. Clearly this works, now we prove it's the only solution. Let $P(x,y)$ denote the given statement. Observe that $P(-1,y)$: $f(y+1) \ge 0 \implies \boxed{f(y) \ge 0 \forall y}$ Now, if $f(y +x^2) = 0$, then taking $x$ positive, $f(y) = 0$, i. e. we have two cases: Case 1: $f(x) \equiv 0$ Which works as discussed above. Case 2: $f(x) = 0 \forall x \le M, f(x) > 0 \forall x > M$. We work on the interval $(M, \infty)$. (Note that $f(x) > 0 \forall x$ is possible, but in this case we perform the same argument on say $(0,\infty)$) Claim 1: $f$ is strictly increasing. Proof: Let $a > b$ i.e. $a = b+m^2$ and take $m$ positive. Now, $f(a) = f(b+k^2) \ge \left(1 + \frac{1}{m} \right)f(b) \ge f(b)$ $\square$ Now say $f(x+\delta) \le kf(x)$ where $k > 1$. Then for sufficiently small $\epsilon$, $$f(x+\delta) > f(x+\epsilon^2) \ge \left(1 + \frac{1}{\epsilon} \right)f(x) > kf(x),$$contradiction. As such, no solutions arise from this case, leading the only solution to be $f\equiv 0$. $\square$

I claim that $f\equiv0$. $P(-1,y)$ : $f(y+1) \ge 0$. So $f$ is non-negative over $\mathbb{R}$. $f$ is increasing : Let $t_{1}$ and $t_{2}$ be real numbers such as $t_1 > t_2$, Then $P( \sqrt{t_1-t_2},t_2)$ : $f(t_1) \ge (\frac{1}{ \sqrt{t_1-t_2}}+1)f(t_2) \ge f(t_2)$. Hence, the result. FTSOC, Let $f(t)>0$ for a fixed real number $t$. $P(x \rightarrow 0^{+}, t)$ : $f(x^{2}+t) \ge (1+ \frac{1}{x})f(t)$ thus the $RHS$ is unbounded as $x$ approaches $0^{+}$. Which contradicts the fact that $f$ is increasing. Hence, $f(x)=0$ for all $x$ in $ \mathbb{R}$. Q.E.D.