Answer: smaller than $1$

Indeed, the key of this problem is to show that $c=1$ fails - simple constructions exist for $c<1$, and proving that no solution exists for $c>1$ is very very easy.

Darn one more time. @rkm0959 Because lack of time, I didn’t check my proving $c=1$ strategy is right. How many point I will lose if I wrote wrong story for $c=1$? I destroyed my afternoon test by myself. Darn!!!!!

Wait, did you write $c=1$ is possible? While I was solving the problem, I was getting the feeling that $c \le 0$ and $c>1$ is worth 1 point, $0<c<1$ is worth 1 point, and $c=1$ is worth 5 points..

rkm0959 wrote: Indeed, the key of this problem is to show that $c=1$ fails - simple constructions exist for $c<1$, and proving that no solution exists for $c>1$ is very very easy. Is $c = 1$ really the hardest case? If $c = 1$, then for any positive integer $n$, \[f(1) = f(\underbrace{n^{-2} + \dots + n^{-2}}_{n^2 \; \text{times}}) \ge f(0) + \underbrace{n^{-1} + \dots + n^{-1}}_{n^2 \; \text{times}} = f(0) + n\]by using $f(x + y^2) \ge f(x) + y$ repeatedly, which is absurd. What am I doing wrong?

Oops sorry It’s impossible.

@CantonMathGuy

I still think $c=1$ is the hardest case because finding the solution is easy for $c<1$ and proving no solution exists for $c>1$ is even easier.

Sorry , but can anyone show the construction for $c <1$? To prove $c\le 1$ is quite easy. We just need to look at the case that $f $ only get negative values. $P (0,x) $ gives contradiction for sufficiently large $x $.

For $c \le 0$, use $f(x)= \sqrt{x}$, and for $c<1$, use $f(x)= \sqrt{\frac{x}{1-c^2}}$. It is easy to prove that these functions $f$ work. Now we show that $c \ge 1$ fails. For $c >1$, plugging $y=0$ gives $f(x) \ge cf(x)$, so $f(x) \le 0$ for all $x$. However, $f(y^2) \ge y+cf(0)$, so taking $y > \text{max}(-cf(0),0)+5000$ gives $f(y^2) \ge 5000$, contradiction. For $c=1$, we show that $f(x) \ge \sqrt{nx}+f(0)$ for all $n \in \mathbb{N}$ and $x \ge 0$, which shows that $f(1)$ cannot exist. We use induction on $n$. For $n=1$, just plug $x=0$ and replace $y$ with $\sqrt{y}$. Assume that it's true for $n$, and we'll show it for $n+1$. Now $f(t) \ge f(x) + \sqrt{t-x} \ge \sqrt{nx} + \sqrt{t-x} + f(0)$ for all $t \ge x$, by setting $t=x+y^2$. Take $x=\frac{nt}{n+1}$, and by simplifying we have $f(t) \ge \sqrt{(n+1)t} + f(0)$, so it's true for $n+1$. This shows that $f(1)$ cannot exist. Done. Therefore, our answer is $c<1$. Note. Here's how I derived the example function $\sqrt{\frac{x}{1-c^2}}$ for $0<c<1$. For $0<c<1$, by plugging $y=0$ we have $f(x) \ge cf(x)$, so $f(x) \ge 0$. This implies that $f(y) \ge \sqrt{y}$, by plugging $x=0$. Denote $a_1=1$ and $a_{n+1}= \sqrt{c^2a^2_n + 1}$. We'll use the same argument we used for $c=1$. By induction, one can show that $f(t) \ge a_n \sqrt{t}$ for all $n \in \mathbb{N}$ and $t \ge 0$. By induction, we can show that $a_n$ is monotonic and $a_n < \sqrt{\frac{1}{1-c^2}}$. By monotonic sequence theorem, we have $\lim_{n \to \infty} a_n = \sqrt{\frac{1}{1-c^2}}$. So $f(x) \ge \sqrt{\frac{x}{1-c^2}}$. Plugging $f(x) = \sqrt{\frac{x}{1-c^2}}$ works, so we're done.

Nice proof @rkm0959 During test, for $c$is smaller than $1$ I just found linear function .

https://artofproblemsolving.com/community/u281710h1543940p9356746 We will prove that $c \ge 1$ are those constants. When $c<1$, one can take $f(x) = x+ \frac{1}{4(1-c)}$. Since \[ f(x+y^2) \ge cf(x) +y \iff (1-c)x \ge y-y^2- \frac{1}{4} \]the condition is satisfied. Now for $c \ge 1$, we will use reduction to the absurdity. Suppose that there exists $f$ with the prescribed domain and codomains which satisfies condition $P(x,y)$ for every $x$ and $y$ in those domain. From $P(0,y)$, we have $f(y^2) \ge cf(0)+y$. Hence we have a positive $k$ such that $x \ge k$ implies $f(x) >0$. Now for every $x \ge k$ and $h \ge 0$, we have $f(x+h)-f(x) \ge (c-1)f(x) + \frac{1}{\sqrt{h}}$. Now for every positive integer $n$, we have \[ f(k+1) - f(k) = \sum_{j=1}^n \left( f \left( k+ \frac{j}{n} \right) - f \left(k + \frac{j-1}{n} \right) \right) \ge \sqrt{n} \]which is an absurdity and completes the proof.

There have been a 2017 KMO(final) in the Contests collection, which is obviously not this one held recently. Does anyone know what happen to the KMO?

jred wrote: There have been a 2017 KMO(final) in the Contests collection, which is obviously not this one held recently. Does anyone know what happen to the KMO? So in order to select Nth year of Korea IMO team, you take first and second round in N-1th year. The second round is usually in November, and about 30~40 students go to winter mop. Then you take some online lecture during spring for MOPers and students take RMM in Feb, APMO in mid-March, and FKMO in late March. Then Korea TST group is decided and they take few more Mock IMO and 6 people are decided in Nth year.

Well, you guys are actually discussing the 2018 KMO(2nd round). Thank you, PARISsaintGERMAIN!

The answer is $c<1.$ Let $P(x,y)$ denote the assertion $$f(x+y^2)\geq cf(x)+y$$We first find that $$P(x,0):\quad f(x)(1-c)\geq 0$$To make our equation simpler let $y=\sqrt{z}$ for $z\in\mathbb{R}_{\geq 0}.$ Then we have \begin{align} P(x,\sqrt{z}):\quad f(x+z)\geq cf(x)+\sqrt{z} \notag \end{align}$i)$ $c>1$ By $P(x,0)$ we can find that $f(x)\leq 0,\forall x\in\mathbb{R}_{\geq 0}.$ $$P(0,\sqrt{z}):\quad f(z)\geq cf(0)+\sqrt{z}$$Let $k$ be an positive real such that $cf(0)+\sqrt{k}>0.$ Then $$P(0,\sqrt{k}):\quad 0\geq f(k)\geq cf(0)+\sqrt{k}>0$$A contradiction. $ii)$ $c=1$ $$P(0,\sqrt{z}):\quad f(z)\geq f(0)+\sqrt{z}$$$$P(z,\sqrt{z}):\quad f(2z)\geq f(z)+\sqrt{z}\geq f(0)+2\sqrt{z}$$Using induction, we can easily show that \begin{align} f(nz)\geq f(0)+n\sqrt{z} \end{align}Plugging in $z\leftarrow\frac{1}{n}$ into (1) we can conclude that $$f(1)-f(0)\geq n\sqrt{\frac{1}{n}}=\sqrt{n}$$which is absurd for appropriately large values of $n.$ Contradiction. $iii)$ $c\leq 0$ $f(x)=\sqrt{x}$ works. $iV)$ $0<c<1$ $$f(x)=\frac{1}{4}x+\frac{1}{1-c}$$works.