For $a \leq 1$ we can find a function $f(x)=kx^2$ such that $3k^2-2k^3=a$ for some $k>0$. Now we will show that for $a>1$ there is no such a function. Since $f$ takes positive values, $f(f(x))>0$ hence $3(f(x))^2>ax^4$ which means $f(x)>\sqrt{\frac{a}{3}}x^2$. Assume that $f(x)>cx^2$ for some constant $c$. Then $f(f(x))>c(f(x))^2>c^3x^4$ and hence $3(f(x))^2>(c^3+a)x^4$ so $f(x)>\sqrt{\frac{c^3+a}{3}}x^2$. Now consider the sequence $y_n$ such that $y_0=\sqrt{\frac{a}{3}}$ and $y_{n+1}=\sqrt{\frac{y_n^3+a}{3}}$ We know that $f(x)>y_nx^2$ for all $x>0$ and non-negative integer $n$. So $y_n$ must be bounded. However, it is easy to see that for $a>1$ the sequence is increasing and has no limit which means it is not bounded. So, we are done.

You have missed $ "2" $ before $ c^3 $ and $ y_n^3 $ so it should be: crazyfehmy wrote: Then $f(f(x))>c(f(x))^2>c^3x^4$ and hence $3(f(x))^2>(2c^3+a)x^4$ so $f(x)>\sqrt{\frac{2c^3+a}{3}}x^2$. Now consider the sequence $y_n$ such that $y_0=\sqrt{\frac{a}{3}}$ and $y_{n+1}=\sqrt{\frac{2y_n^3+a}{3}}$ We know that $f(x)>y_nx^2$ for all $x>0$ and non-negative integer $n$. So $y_n$ must be bounded. However, it is easy to see that for $a>1$ the sequence is increasing and has no limit which means it is not bounded. So, we are done.

kucheto wrote: You have missed $ "2" $ before $ c^3 $ and $ y_n^3 $ so it should be: crazyfehmy wrote: Then $f(f(x))>c(f(x))^2>c^3x^4$ and hence $3(f(x))^2>(2c^3+a)x^4$ so $f(x)>\sqrt{\frac{2c^3+a}{3}}x^2$. Now consider the sequence $y_n$ such that $y_0=\sqrt{\frac{a}{3}}$ and $y_{n+1}=\sqrt{\frac{2y_n^3+a}{3}}$ We know that $f(x)>y_nx^2$ for all $x>0$ and non-negative integer $n$. So $y_n$ must be bounded. However, it is easy to see that for $a>1$ the sequence is increasing and has no limit which means it is not bounded. So, we are done. Yes, you are right, thanks.

crazyfehmy wrote: For $a \leq 1$ we can find a function $f(x)=kx^2$ such that $3k^2-2k^3=a$ for some $k>0$. Now we will show that for $a>1$ there is no such a function. Since $f$ takes positive values, $f(f(x))>0$ hence $3(f(x))^2>ax^4$ which means $f(x)>\sqrt{\frac{a}{3}}x^2$. Assume that $f(x)>cx^2$ for some constant $c$. Then $f(f(x))>c(f(x))^2>c^3x^4$ and hence $3(f(x))^2>(c^3+a)x^4$ so $f(x)>\sqrt{\frac{c^3+a}{3}}x^2$. Now consider the sequence $y_n$ such that $y_0=\sqrt{\frac{a}{3}}$ and $y_{n+1}=\sqrt{\frac{y_n^3+a}{3}}$ We know that $f(x)>y_nx^2$ for all $x>0$ and non-negative integer $n$. So $y_n$ must be bounded. However, it is easy to see that for $a>1$ the sequence is increasing and has no limit which means it is not bounded. So, we are done. I don't understand.

just Think about the sequence that a_i=f(f(...(f(x))). we can get inequailty when a>1.

We claim the answer is all $a\le 1$. Note that the image of \[g(x) = 3x^2-2x^3\]over the positive reals is the set $(-\infty, 1]$, so, for those values of $a,$ take $d$ such that \[3d^2-2d^3 = a,\]and $f(x) = d.x^2$ works. Now suppose $a>1$. Let $(y_n)_n$ be the sequence of non-negative real numbers defined by $y_0=0$, \[y_{n+1}^2 = \frac{2y_n^3+a}{3}\] Claim: $y_n>1$ for all sufficiently large $n$. Proof: We first prove it is bigger than one for some $n$ and it follows from induction. Suppose $y_n\le 1$ for all $n$. Then, if $b_n = y_n^3 - a$, \[b_{n+1} \le b_n.\frac{2}{3},\]which implies $b_n$ goes to zero, which implies $y_n$ tends to $\sqrt[3]{a}$, which is greater than one; a contradiction. This proves the claim. Note that, by induction on the original equation, $f(x)>y_nx^2$, so $(y_n)$ is bounded and \[y_{n+1}^2 - y_n^2 = \frac{2y_n^3-3y_n^2+1}{3}>0,\]so it converges to some $L$. But this implies that $g(L) = 3L^2-2L^3 = a$, contradicting $a>1$. $\blacksquare$