Bashing is always our friend.

I wonder where you use the second condition

FEcreater wrote: other given condition tells us that $ g_y\left(x\right) $ is bounded(Try to verify it!)

Edit:I think this is similar to FEcreater's solution.

bgn wrote: Find all functions $f: \mathbb {R}^+ \times \mathbb {R}^+ \to \mathbb {R}^+$ that satisfy the following conditions for all positive real numbers $x,y,z:$ $$f\left ( f(x,y),z \right )=x^2y^2f(x,z)$$$$f\left ( x,1+f(x,y) \right ) \ge x^2 + xyf(x,x)$$ Proposed by Mojtaba Zare, Ali Daei Nabi Hope this works. This problem has been bothering me since last May. First we vary $y$ so for fixed $z$, we see $f(--,z)$ is surjective. Suppose for fixed $z>0$ we have $f(a,z)=f(b,z)$; then $$a^2z^2f(a,z)=f(f(a,z),z)=f(f(b,z),z)=b^2z^2f(b,z)$$proving $a=b$ and establishing injectivity of $f(--,z)$. Plug $x=y=1$ and apply injectivity so $f(1,1)=1$. Compute $f(f(f(a,b),c),d)$ in two ways: $$f(f(f(a,b),c),d)=f(a,b)^2c^2a^2b^2f(a,d)$$and $$f(f(f(a,b),c),d)=f(a^2b^2f(a,c),d)$$for all $a,b,c,d>0$. Plug $abcf(a,b)=1$ for a good choice of $c$; then injectivity gives $f(a,c)=\frac{1}{ab^2}$. Therefore, plugging $x=a, y=c, z$ yields $$f\left(\frac{1}{ab^2}, z\right)=\frac{f(a,z)}{b^2f(a,b)^2}$$for all $z>0$. Now comes the intent we had while using this crazy choice of $c$: put $sab^2=1$ and $a=t$; then $\frac{f(s,z)}{f(t,z)}$ depends only on $s,t$. Hence we can write $$\frac{f(x,y)}{f(1,y)}=\frac{f(x,1)}{f(1,1)} \implies f(x,y)=f(x,1)f(1,y)$$for all $x,y>0$. In the original, plug $x=z=1$ so $f(f(1,y),1)=y^2$ hence $$f(y^2,1)=f(f(f(1,y),1),1)=y^2f(1,y)^2$$for all $y>0$. Thus, we get $$f(a,1)f\left(\frac{1}{ab^2},1\right)=f\left(\frac{1}{b^2},1\right)$$for all $a,b>0$ thus proving $f(x,1)f(y,1)=f(xy,1)$ for all $x,y>0$. Let $g(x)=f(x,1)$ then $g$ maps positive reals to themselves with $g(xy)=g(x)g(y)$ for all $x,y>0$. Note also that condition 2 is just that $g$ has a lower bound on some interval. Hence $g(x)=x^k$ for some $k \in \mathbb{R}$. Now checking gives $k=2$ and $f(1,y)=y$ for all $y>0$. Thus, $f(x,y)=x^2y$ for all $x,y>0$. This clearly works.

bgn wrote: Find all functions $f: \mathbb {R}^+ \times \mathbb {R}^+ \to \mathbb {R}^+$ that satisfy the following conditions for all positive real numbers $x,y,z:$ $$f\left ( f(x,y),z \right )=x^2y^2f(x,z)$$$$f\left ( x,1+f(x,y) \right ) \ge x^2 + xyf(x,x)$$ Proposed by Mojtaba Zare, Ali Daei Nabi Let $P(x,y,z)$ be assertion $f(f(x,y),z)=x^2y^2f(x,z)$ Let $f(x,1)=g(x)$ and $f(1,y) = h(y)$ \begin{align} &P(1,y,1) \implies g(h(y)) = y^2g(1) \\ &P(x,1,1) \implies g(g(x)) = x^2g(x) \end{align}$(1)$ and $(2)$ mean $g$ is bijective. Which in turn will imply $h$ is bijective. Also it is easy to see from here $f(1,1)=h(1)=g(1)=1$. \begin{align*} P(1,y,z) &\implies f(h(y),z) = y^2h(z) \\ &\implies f(h(y),z) = f(h(y),1)f(1,z) \\ &\implies f(h(y),z) = g(h(y)) h(z) \\ \end{align*}Since '$h$' is surjective, setting $h(y) = x$ $$ \boxed{\implies f(x,y) = g(x)h(y)}$$Replacing this in the original equation, and using $(1)$ and $(2)$, \begin{align*} &P(x,y,z) \equiv g(g(x)h(y))=x^2y^2g(x) \\ &\implies g(g(x)h(y)) = g(g(x))g(h(y)) \\ \end{align*}Now since $g$ and $h$ are surjective, $$\implies \boxed{g(xy)=g(x)g(y)}$$ Now the second condition on the question implies, $g$ is bounded in an interval. Hence by using Cauchy equation, we easily obtain, $$g(x)=x^2$$And thereby showing $\boxed {f(x,y) = x^2y}$

utkarshgupta wrote: Now the second condition on the question implies, $g$ is bounded in an interval. How does the condition prove $g$ is bounded?

TLP.39 wrote:

could you tell me why？

Similar to above solutions, reminded me of 2019 TSTST/1.

Here is another mass solution; collaborators include Aahan Chatterjee, Aditya Khurmi, Alexandru Girban, Anushka Aggarwal, Derek Liu, Eric Shen (CAN), Maximus Lu, Paul Hamrick, Robin Son, Sean Li, William Wang, Zipeng Lin. The answer is $f(x,y) = x^2y$ only which works. We will use only the first equation for most of this solution, only invoking the inequality at the very end. Let $P(x,y,z)$ be the given assertion. Claim: The first equation implies (actually is equivalent to) having \[ f(x,y) = g(x) \cdot \frac{g(y)}{y} \]for some completely multiplicative function $g$. Proof. First we will try to show $f(x,y)=g(x)h(y)$. Define the function $\boxed{g(x)=f(x,1)}$. Then $P(x,1,1)$ gives \[ g(g(x)) = x^2 g(x) \qquad(\heartsuit) \]so $g$ is injective. Note $P(1,1,1)$ means $g(g(1)) = g(1)$; as $g$ is injective we get $g(1) = 1$. Define $\boxed{h(y)=f(1,y)}$. Considering $P(1,y,1)$ gives \[ g(h(y)) = y^2 \qquad (\spadesuit) \]which implies $g$ is bijective, and thus $h$ is bijective too. For any $a,b > 0$, note that $P(1,h^{-1}(a),b)$ gives \[ f(a, b) = a^2 h(b) \overset{(\heartsuit)}{=} g(a) h(b). \] We can then rewrite the equation in terms of $g$ and $h$: we have $g(g(x)h(y))h(z) = x^2 y^2 g(x) h(z)$ and so by $(\heartsuit)$ and $(\spadesuit)$ we get \[ g(g(x)h(y)) = x^2g(x) \cdot y^2 = g(g(x)) \cdot g(h(y)) \]which, since $g$ and $h$ are bijections, means $g$ is completely multiplicative. Finally, if we put $x=h(t)$ in $(\heartsuit)$ then we get $g(t)^2 = h(t)^2 \cdot t^2$ which means $h(t) = g(t)/t$, finishing the claim. $\blacksquare$ We finally use the inequality. Fix $x > 1$. By varying $y$, the number $a = 1+f(x,y) = 1 + g(x)h(y)$ may take on any value (since we saw $h$ was surjective). Then \[ g(x) \cdot \frac{g(a)}{a} > x^2 \iff g(ax) > ax^2 > x^2. \]And in conclusion we find (among other things) that $g(x) > 1$ for all $x > 1$. This implies that $t \mapsto \log g(e^t)$ is an additive function $\mathbb R \to \mathbb R$ which is bounded below on $(0, \infty)$; so Cauchy implies it is linear. Thus $g(x) = x^k$ for some constant $k$. It is routine to check now that we must have $k=2$ which gives our solution.

here's a shorter solution $P(1,1,z) \implies f(f(1,1),z)=f(1,z)$ combining with $P(f(1,1),1,z) \implies f(1,1)=1$ Now put $g(x)=f(x,1)$ $h(x)=f(1,x)$ $P(1,x,1) \implies g(h(x))=x^2 ...... (1)$ $P(x,1,1) \implies g(g(x))=x^2h(x)^2 .....(2)$ from (1) and (2) we can conclude $g(x^2)=x^2h(x)^2 .....(3) $ $h(xh(x))=x^2 .....(4) $ $P(1,y,z) \implies f(h(y),z)=y^2h(z)$ put $y =yh(y)$ and from (3) and (4) we will have $f(x,y)=g(x)h(y)$ $P(x,y,1) \implies g(g(x)h(y))=x^2g(x)y^2=g(g(x)),g(h(y)) \implies g(ab)=g(a)g(b)$ so from (1) $g(x)=xh(x)$ from (4) $h(h(x))=x$ and of course $h(ab)=h(a)h(b)$ from the second inequality we will have: $h(x)h(1+xh(xy) \ge x+xyh(x)^2 $ put $y=xy $ $h(x)h(1+xh(y) \ge x+yh(x)^2 $ put $x=h(x)$ $h(x) \ge x $ but $x=h(h(x)) \ge h(x)$ thus $h(x)=x \implies$ $f(x,y)=x^2y$ and we win

idk this seemed a lot harder last year. .

The first assertion is very strong. We can deduce a lot from it, and the second one gives some analytical bounds to finish the problem. The answer is $f(x,y)\equiv x^2y$, which clearly works. Let $P(x,y,z), Q(x,y)$ denote the first and second assertion in the problem. Claim 1: If $f(a,z)=f(b,z)$ for some $a,b,z\in \mathbb{R}^+$ then $a=b$. Proof: $P(a,z,z), P(b,z,z)$ yields $a^2z^2f(a,z)=f(f(a,z),z)=f(f(b,z),z)=b^2z^2f(b,z)$ Since $f(a,z)=f(b,z)$, $a^2z^2=b^2z^2 \rightarrow a=b$ Now, $P(x, \frac 1x, z)$ yields $f(f(x,\frac 1x),z)=f(x,z)$, so $f(x,\frac 1x)=x$ $P(x,y,\frac 1x)$ yields $f(f(x,y),\frac 1x)=x^3y^2$ Now, we consider $$f(f(f(x,y),z) ,w)$$. On one hand, by $P(f(x,y),z,w)$, it is equal to $f(f(x,y),w) f(x,y)^2 z^2=(xyzf(x,y))^2 f(x,w)$ If $xyzf(x,y)=1 \iff z=\frac{1}{xyf(x,y)}$ then by injectivity we get $$f\left(f(x,y), \frac{1}{xyf(x,y)}\right)=x$$ $$f\left( x, \frac{1}{xyf(x,y)}\right) = \frac{1}{xy^2}$$ By $P(x, \frac{1}{xyf(x,y)}, z)$, we obtain $$\frac{f\left( \frac{1}{xy^2}, z\right)}{f(x,z)}=\frac{1}{y^2 f(x,y)^2}$$ This implies $$\frac{f(u,z)}{f(v,z)}$$doesn't depend on $z$, so there exists functions $g,h$ such that $f(x,y)\equiv g(x)h(y)$ Wlog $g(1)=1$, then note by substituting $z=y$ in the above equation we have $$f\left( \frac{1}{xy^2}, y\right) = \frac{1}{y^2 f(x,y)}$$ $$g\left( \frac{1}{xy^2} \right) h(y) = \frac{1}{y^2 g(x)h(y)}$$ $$g\left(\frac{1}{xy^2}\right) g(x) = \frac{1}{y^2h(y)^2}=g(1)g\left( \frac{1}{y^2} \right)$$ We can get $g(x)g(y)=g(xy)$ for all $x,y\in \mathbb{R}^+$ and $g(x)=xh(x)$. Note $h(x)h(y)=h(xy)$ as well. Now, we expand $Q(x,y) \colon g(x) h(1+g(x)h(y)) \ge x+xyg(x)h(x)$ $$h(x) h(1+xh(xy))\ge x+xyh(x)^2$$ Let $t=xy$ then $h(x)h(1+xh(t))\ge x^2+th(x)^2$ for all $x,t\in \mathbb{R}^+$. I claim $h(x)\ge 1$ for all $x\ge 1$. Suppose $h(t)<1$, then $x=\frac{1}{1-h(t)} \to $ $1+xh(t)=x$, so $h(x)^2\ge x+th(x)^2$, so $t<1$ Now, note the function $k(x)=\log(h(e^x))$ is additive and $k(x)>0 \forall x>0$. It follows that $k(x)\equiv cx \rightarrow h(x)\equiv x^d$ for some $d>1$. This implies $h(x)=x, g(x)=x^2$. The end.

Let $P(x,y,z)$ be the assertion of the first condition , so one can easily see that $f(x,y)$ is injective on each vertical and horizontal line ( means that for each numbers $x , y , z \in \mathbb{R^+}$ , we have $f(x,z)=f(y,z) \iff x=y$ and $f(z,x)=f(z,y) \iff x=y$). As the result , we can get : $$P(x , \frac{1}{x} , z) \implies f(f(x , \frac{1}{x}) , z)=f(x,z) \implies f(x , \frac{1}{x})=x (I)$$$$P(x , \frac{\sqrt{f(t,z)}}{x\sqrt{f(x,z)}} , z) \implies f(f(x,\frac{\sqrt{f(t,z)}}{x\sqrt{f(x,z)}}),z)=f(t,z) \implies f(x , \frac{\sqrt{f(t,z)}}{x\sqrt{f(x,z)}})=t \implies \frac{f(t,z)}{f(x,z)}=\frac{f(t,y)}{f(x,y)} , \forall x , y , z , t \in \mathbb{R^{+}} (II) $$Now putting $t=z=1$ in the equation $(II)$ , since we know $f(1,1)=1$ one can see that we have $f(x,y)=f(x,1)f(1,y)$ for all positive real numbers $x,y$. So if we define new functions $g , h: \mathbb{R^+} \to \mathbb{R^+}$ such that $f(x,1)=g(x)$ , $f(1,x)=h(x)$ for all numbers $x \in \mathbb{R^+}$ , then $f(x,y)=g(x)h(y)$ and concerning equation $(I)$ we have $g(x)h(\frac{1}{x})=x$ and one can see that : $$f(f(x,y),1)=x^2y^2f(x,1) \implies g(g(x)h(y))=x^2y^2g(x) \implies g(\frac{yg(x)}{g(y)})=\frac{x^2}{y^2}g(x) , \forall x , y \in \mathbb{R^+}$$Now if we show this new assertion as $Q(x,y)$ , by $Q(x,1)$ we can get $g(g(x))=x^2g(x)$ and since obviously $g(x)$ is a surjective function on positive real numbers , we can get : $$g(\frac{yg(x)}{g(y)})=\frac{g(g(x))}{y^2} \implies g(\frac{xy}{g(y)})=\frac{g(x)}{y^2} \implies Q(x , g(y)) : g(\frac{x}{y^2})=\frac{g(x)}{g(y)^2} (S(x,y))$$$$S(y^2 , y) \implies g(y^2)=g(y)^2 \implies g(xy)=g(x)g(y) \forall x,y \in \mathbb{R^+}$$So while $g(x)$ is multiplicative and bounded on some interval by second condition ( $g(x) \ge 1$ for all $x \ge 1$ ) , as the result there exist a real number $a \in \mathbb{R}$ such that $g(x)=x^a$ for all positive real numbers $x$ and since $g(g(x))=x^2g(x)$ , we can get $g(x)=x^2$ and $h(x)=x$ , so we have $f(x,y)=x^2y$ for all positive real numbers $x , y$ and we're done.