Saudi Arabian IMO Booklet

1) In the given condition, replace $x \rightarrow 3x, y \rightarrow 2y$, we have
$$f(3x + 4y + f(3x + 2y)) = f(6x) + f(6y).$$
By swapping $x, y$ and comparing the two left sides, we have
$$f(3x + 4y + f(3x + 2y)) = f(4x + 3y + f(2x + 3y)).$$
From injectivity, we have
$$3x + 4y + f(3x + 2y) = 4x + 3y + f(2x + 3y)$$
or
$$f(3x + 2y) - (3x + 2y) = f(2x + 3y) - (2x + 3y).$$
Set $g(x) = f(x) - x$ then $g(3x + 2y) = g(2x + 3y), \forall x, y \in \mathbb{R}^+$. Put $a = 3x + 2y, b = 2x + 3y$ then
$$x = \frac{3a - 2b}{5}, \quad y = \frac{3b - 2a}{5}$$
so the condition of the pair of numbers $(a, b)$ for the existence of $x, y$ the above relation is $3a - 2b > 0$ and $3b - 2a > 0$, equivalent to $\frac{2}{3} < \frac{a}{b} < \frac{3}{2}$.
Hence we have $g(a) = g(b)$ for all $\frac{a}{b} \in (\frac{2}{3}; \frac{3}{2})$. From here we will show that $g$ is a constant function. Indeed, we have $g(1) = g(x), \forall x \in [1; \frac{3}{2})$, from here inductively
$$g(1) = g(x), \forall x \in \left[ \left(\frac{3}{2}\right)^k ; \left(\frac{3}{2}\right)^{k+1} \right)$$
so $g(1) = g(x), \forall x > 1$. Similarly
$$g(1) = g(x), \forall x \in \left( \left( \frac{2}{3} \right)^{k+1} ; \left( \frac{2}{3} \right)^k \right]$$
so there is also $g(1) = g(x), \forall x \in (0; 1)$. Therefore, $g(x)$ is a constant function. Instead we have $f(x) = x + c$ with $c \geq 0$, try again and we are satisfied.

2) Since $f(x) = x + c$, instead of the problem condition, we have
$$(4\sin^4x + c)(4\cos^4x + c) \geq (1 + c)^2 \text{ or } \sin^4(2x) + 4(\sin^4x + \cos^4x)c \geq 1 + 2c.$$
Notice that $\sin^4x + \cos^4x = 1 - \frac{1}{2}\sin^22x$ and put $t = \sin^2(2x) \in (0; 1]$, we rewrite the above inequality as $t^4 + (4 - 2t^2)c \geq 1 + 2c$ or
$$2(1 - t^2)c \geq (1 - t^2)(t^2 + 1).$$
Since $1 - t^2 > 0$ so $2c \geq t^2 + 1$, which $\max_{(0;1]}\{t^2 + 1\} = 2$ so $2c \geq 2 \Leftrightarrow c \geq 1$. So the minimum value of $f(2022)$ is 2023. □