Saudi Arabian IMO Booklet

1) In the given condition, replace $x\to 3x,y\to 2y$, 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 $$x=\frac{3a-2b}{5},y=\frac{3b-2a}{5}$$ 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 $c\ge 0$, try again and we are satisfied.

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