fun problem! but a little easy for a TST I think.. Mr.C wrote: Let $f : \mathbb{R}^+\to\mathbb{R}$ satisfying $f(x)=f(x+2)+2f(x^2+2x)$. Prove that if for all $x>1400^{2021}$, $xf(x) \le 2021$, then $xf(x) \le 2021$ for all $x \in \mathbb {R}^+$ Proposed by Navid Safaei Let $g(x)=\frac{2021}{x}-f(x)$. Then $g(x) \ge 0$ for all $x>1400^{2021}$. since $\frac{1}{x}=\frac{1}{x+2}+2*\frac{1}{x^2+2x}$ , $\frac{2021}{x}=\frac{2021}{x+2}+2*\frac{2021}{x^2+2x}$. So, $g(x)=g(x+2)+2g(x^2+2x)$ because $x+2, x^2+2x>x$ , If we repeat the progress of expressing $g(x)$ into $g(x+2), g(x^2+2x)$ , we can get $g(x) \ge 0$ for all $x \in \mathbb {R}^+$

doesnt seem that your solution is complete

I claim that if $xf(x)\leq 2021$ for all $x>c$ then it also holds for all $x$ in the interval $(c-2,c]$ and with it continue inductively in $\mathbb {R}^+$. $f(x)=f(x+2)+2f(x^2+2x)\leq \tfrac {2021}{x+2}+2\tfrac {2021}{x(x+2)}=\tfrac {2021}{x}$.

Com10atorics wrote: I claim that if $xf(x)\leq 2021$ for all $x>c$ then it also holds for all $x$ in the interval $(c-2,c]$ and with it continue inductively in $\mathbb {R}^+$. $f(x)=f(x+2)+f(x^2+2x)\leq \tfrac {2021}{x+2}+\tfrac {2021}{x(x+2)}=\tfrac {2021}{x}$. only works when $c \ge3 $

Mr.C wrote: Let $f : \mathbb{R}^+\to\mathbb{R}$ satisfying $f(x)=f(x+2)+2f(x^2+2x)$. Prove that if for all $x>1400^{2021}$, $xf(x) \le 2021$, then $xf(x) \le 2021$ for all $x \in \mathbb {R}^+$ Proposed by Navid Safaei Hopefully didn't miss any stupid details. Claim 01. $xf(x) \le 2021$ for all $x \ge 1$. Proof. To prove this. Let $xf(x) \le 2021$ for all $x > c$, where $c \ge 3$. Take $x = c - 2 + \varepsilon$ to the equation above, and check that \[ x + 2 = c + \varepsilon > c \ \text{and} \ x(x + 2) = (c - 2 + \varepsilon)(c + \varepsilon) > c + \varepsilon > c \]Therefore, we conclude that \begin{align*} xf(x) &= xf(x + 2) + 2xf(x^2 + 2x) \\ &\le x \left( \frac{2021}{x + 2} \right) + 2x \left( \frac{2021}{x^2 + 2x} \right) \\ &= 2021. \end{align*}Hence, if the inequality is true for some $x > c$, $c \ge 3$, then it is also true for $x > c - 2$. Furthermore, for $x = 1$, we have $f(1) = 3f(3) \le 2021$, as desired. Claim 02. $xf(x) \le 2021$ for all $x \in (0,1)$. Proof. Suppose that we have $xf(x) \le 2021$ for all $x > c, c \in (0,3)$. Taking $x = \sqrt{c + 1} - 1 + \varepsilon$, which is positive, we check that \[ x + 2 = \sqrt{c + 1} + 1 + \varepsilon > \sqrt{c + 1} + 1 > c \]\[ x^2 + 2x = (x + 1)^2 - 1 = (\sqrt{c + 1} + \varepsilon)^2 - 1 > (\sqrt{c + 1})^2 - 1 = c \]Therefore, we conclude that $xf(x) \le 2021$ for all $x > \sqrt{c + 1} - 1$. We define $\alpha$ to be fun if for all $x > \alpha$, then $xf(x) \le 2021$, we have proved that if $c \in (0,3)$ is fun, then $\sqrt{c + 1} - 1$ is fun as well. Now, define a sequence $a_0 = c > 0$, and $a_{i + 1} = \sqrt{a_i + 1} - 1$. We claim that $\lim_{i \to \infty} a_i = 0$. To prove this, just notice that \[ \lim_{i \to \infty} a_i = \lim_{i \to \infty} (a_0 + 1)^{\frac{1}{2^i}} - 1 = 0 \]Therefore, we have proved that $xf(x) \le 2021$ for all $x \in \mathbb{R}^+$.