$P(1,1) \implies f(1)^2+2f(1)+1 \geq 3f(1)^2+1 \implies 1 \geq f(1) \implies f(1)=1$ $P(1,n) \implies (f(n)+1)^2>f(n)^2+2f(n) \geq n^2+2$ If $f(n) \leq n-1$ then $n^2 \geq n^2+2$ Contradiction So $f(n) \geq n$ $P(m,1) \implies m^2+2m \geq 3f(m)^2 \geq 3m^2 \implies 1 \geq m$ Contradiction

Solution (Walkthrough): TSTs going to be so easy, unbelievable ! Put $m=n=1$ to get $$f(1)^2-f(1) \leq 0 \Longrightarrow f(1)=1$$Next put $n=1$ to get $$m^2+2m \geq 3f(m)^2$$Again put $m=1$, to get $$n^2+2 \leq f(n)^2+2f(n)$$which is only possible when $f(n) \geq n$. Compare it to the first one and since the function is defined in between natural numbers, we will get no functions exist.