For any $(a,b)\in\mathbb R^2$ such that $2f(a)+f(-b)\geq 0$, we have: $2f(-a)+f(b)\leq 0$. Then, $f$ must be an odd function?

Question

For any $(a,b)\in\mathbb R^2$ such that $2f(a)+f(-b)=0$, we have: $2f(-a)+f(b)=0$.

For any $(a,b)\in\mathbb R^2$ such that $2f(a)+f(-b)>0$, we have: $2f(-a)+f(b)<0$.

Also, for all $a$, there exists $b$ such that $2f(a)+f(-b)=0$.

$f:\mathbb R\to\mathbb R$ is continuous. $f(x)\geq0$ for all $x\geq 0$. $f(x)\leq0$ for all $x\leq0$.

Based on the above conditions, can we prove or disprove that $f$ is an odd function: $f(x)=-f(-x)$?

The first condition means that: if $2f(2)+f(-1)=0$, then $2f(-2)+f(1)=0$.

I think we cannot prove this. Yet I failed to find any counterexamples against the claim.

Let me know if the logic of the question is ambiguous or clear.