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Please, can someone post a solution for this? Thanks!

the only thing I could proove is that $f(x)=f(\frac{1}{x})$...and actually something like $f(x)+f(y)=f(\sqrt{\frac{x}{y}})f(\sqrt{xy})$ my hipothesis is $f(x)=x^m+\frac{1}{x^m}$ , but i still can't get it ... it's difficult anyway ...

It can't be that hard. The desired answer is indeed $f_m(x)=x^m+\frac{1}{x^m}$ , for any $m>0$. It is easy to check that it satisfies our equation. I think I have an "ugly solution" for this. Fix some $t>1$ as given and $f(t)=t^m+\frac{1}{t^m}=r>2$ , $r$-given (from this it follows some $m$). We can express $f(t^2)=f^2(t)-2$ , $f(t^3)=f^3(t)-3f(t)$ , ... and generally $f(t^n)=P_n(f(t))$, for any integer $n>0$ , where $P_n$ are polynomials ($n$-even, $P_n$-even ; while $n$-odd, $P_n$-odd and coefficients in $P_n$ are only function of $n$ - I computed them recursively and they have an "ugly" form in my oppinion). Hence $f(t^n)$ just follows uniquely determined from the given $f(t)$ - anyway i don't need those coefficients in explicitely form, since my goal is to show that $r=P_n(x)$ has only one solution greater than $2$, $x>2$ for $r>2$ - it results by induction this important fact in the proof; here it is used the increasing condition. The equation $f(t)=P_n(f(t^{1/n}))$ has an unique solution for $f(t^{1/n})>2$ uniquely determined by that initial value $f(t)=r>2$. So for any rational $p/q>0$ I can compute uniquely $f(t^{p/q})$ from $f(t^p)$ which come from $f(t)$, then my function is well-determined for any rational $x=t^{p/q}>1$ as $f(x)$ starting from that $f(t)=r>2$, then it follows by extension for any real $x>1$, using again increasing condition! Since $f(x)=f(\frac{1}{x})$ we get the only possible answer $f_m(x)=x^m+\frac{1}{x^m}$ , $f_m: \mathbb{R}^+ \longrightarrow \mathbb{R}^+$ for any $m>0$. I hope that some other "good looking" proof exists.

this solution doesn't convince me , I'll will have to look through this concise deduction again ....

Which part of the proof you don't agree?

heartwork wrote: ... The desired answer is indeed $f_m(x)=x^m+\frac{1}{x^m}$ , for any $m>0$. Fix some $t>1$ as given and $f(t)=t^m+\frac{1}{t^m}=r>2$ , $r$-given (from this it follows some $m$). We can express $f(t^2)=f^2(t)-2$ , $f(t^3)=f^3(t)-3f(t)$ , ... and generally $f(t^n)=P_n(f(t))$, for any integer $n>0$ , where $P_n$ are polynomials ... I have seen both official solutions for this problem! Surprisingly, both uglier and much longer than mine! These polynomials obey a similar recurrence as the Chebyshev polynomial of degree n: $P_n(2x)$ similar to $p_n(x) = $ cos$(n $ arccos$(x)) $ $p_{n+1}(x) = 2xp_n(x) - p_{n-1}(x)$

Lemma 1: $ f(x) \geq 2 $ for all $ x \in \mathbb{R}^{+} $. Proof: Note that by letting $ x = y = z = 1 $ in the equation we find that $ f(1) = 2 $ and then letting $ y = z = 1 $ in the equation and replacing $ x $ by $ x^2 $ we have that $ f(x^2) = f(x)^2 - 2 $. This implies $ f(x) > \sqrt{2} $ for all $ x \in \mathbb{R}^{+} $. Using this result we now get $ f(x) > \sqrt{2+\sqrt{2}} $ for all $ x \in \mathbb{R}^{+} $. Continuing in this fashion we obtain the desired result. Lemma 2: Let $ n $ be a nonnegative integer. Let $ p_{n} $ be a polynomial defined by the following criteria: $ p_0 = 2, p_1 = x, p_{k + 1} = xp_k - p_{k - 1} $ for all positive integers $ k $. Then $ f(t^n) = p_n(f(t)) $ for all $ t \in \mathbb{R}^{+} $. Proof: It is clear that $ p_n\left(t + \frac{1}{t}\right) = t^n + \frac{1}{t^n} $ for all real $ t $ (this is easily proven by induction) so it suffices to show that $ f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1}) $ and $ f(t^{2n}) = f(t^n)^2 - 2 $. We already know that $ f(x^2) = f(x)^2 - 2 $ for all $ x \in \mathbb{R}^{+} $ so the second identity follows immediately. To prove the first identity, let $ x = t^{2n -1 } $ and $ y = z = t $ in out equation. Then we obtain $ f(t^{2n + 1}) + f(t^{2n - 1}) + 2f(t) = f(t)f(t^{n})^2 = f(t)(f(t^{2n}) + 2) $ and so $ f(t^{2n + 1}) = f(t)f(t^{2n}) - f(t^{2n - 1}) $ as desired. Lemma 1 implies that for any $ s \in \mathbb{R}^{+} $ we can write $ f(s) = a + \frac{1}{a} $ for some real $ a \geq 1 $. Lemma 2 then implies that $ f(s^q) = a^q + \frac{1}{a^q} $ for all $ q \in \mathbb{Q} $. Because the rationals are dense in the reals and because $ f $ is increasing on the interval $ [1, \infty) $ this implies that $ f(s^r) = a^r + \frac{1}{a^r} $ for all $ r \in \mathbb{R} $. This clearly implies that $ f(x) = x^m + \frac{1}{x^m} $ for some $ m \in \mathbb{R} $ and it is easy to check that all such functions satisfy the original equation.

Can sb explain the solution here?(problem 14) http://www.imomath.com/index.php?options=341&lmm=0

bump!!!!!!!!!!!!!!!!

Finally solved this, yay! (I had tried this problem around 8 months back, but had failed to solve it then.) orl wrote: Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy the following conditions: - $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$; - $f(x)<f(y)$ for all $1\le x<y$. Proposed by Hojoo Lee, Korea Answer : $f(x) = x^k + \frac{1}{x^k}$ for some $k > 0$. One can confirm that this function does work. $x=y=z=1 \implies 4 \cdot f(1) = f(1)^3 \iff f(1) = 2$. Now $x^2,1,1$ gives us that $$ 2f(x^2) + 4 = 2f(x)^2 \implies f(x^2) + 2 = f(x)^2 \dots (\star)$$ Claim: $f(x) = h(x) + \frac{1}{h(x)}$ for some function $h : \mathbb{R^+} \rightarrow \mathbb{R^+}$. $ $ Proof. From $(\star)$ we can guarantee that $f \ge \sqrt{2}$. But this again gives us that $f \ge \sqrt{2+\sqrt{2}}$. Going on recursively, we have that $f \ge 2$. Hence such a function $h$ exists. Moreover define $h(x) \ge 1$ whenever $x \ge 1$. So the property of $f$ asserts that $h$ is increasing on $(1 , \infty)$. Also note that such a function $h$ nicely interprets the condition $\star$. $ $ Now consider $Q\left(x,x,\frac{1}{x}\right)$ to get that $f(x) = f\left(\frac{1}{x}\right) \implies h\left(\frac{1}{x}\right) = \frac{1}{h(x)} \ \forall x \implies h$ is increasing on $(0, \infty)$. Using $(\star)$ on $Q(x^2,y^2,1)$ we also get that $$ f(x)^2 + f(y)^2 + f(xy)^2 = f(x)\cdot f(y) \cdot f(xy) + 4$$This is indeed a very famous setup. We get that there exist positive reals $a,b,c$ such that $abc=1$ and $f(x) = a + \frac{1}{a}$ and so on. Thus, using these substitutions we arrive at the fact that $h$ is multiplicative and considering the function $g = \log{f(a^x)}$ we get that $g$ is additive and increasing $\implies g(x) = cx \implies h(x) = x^k$ for some $k$.(This follows from $h(1)=1$). Hence we have our solution.

We claim that the solutions are $x^m + 1/x^m$ where $m$ is any positive real number. It's easy to see these work, so we'll prove these are the only solutions. Let $P(x, y, z)$ denote the condition $f(x^2y^2z^2) + f(x^2) + f(y^2) + f(z^2) = f(xy)f(yz)f(zx)$ for all positive reals $x, y, z$ (equivalent to the first condition of the problem). $P(1, 1, 1) \implies f(1) = 2$. $P(z, 1, 1) \implies f(z^2) = f(z)^2-2$. $P(z, z, 1/z) \implies f(z) = f(1/z)$. Subtracting $P(x, y, z)$ from $P(x, 1/y, 1/z)$ gives $f(x^2/(y^2z^2)) - f(x^2y^2z^2) = f(x/y)f(x/z)f(yz) - f(xy)f(xz)f(yz)$. Putting $y=1$ here gives $f(x^2/z^2) - f(x^2z^2) = f(x)f(z)(f(x/z) - f(xz))$. So $f(x/z) + f(xz) = f(x)f(z)$ ($f(x/z) \neq f(xz)$ if neither of $x, z$ equals $1$ (easily seen by the second condition of the problem and $f(a) = f(1/a)$). The case where one of them equals $1$ follows independently). Let $f(2) = 2^a + 1/2^a$ where $a$ is a positive real number (note that the function $2^x + 1/2^x$ for $x > 0$ has range $(2, \infty)$ and is strictly increasing). Call a positive real number $k$ good if $f(k) = k^a + 1/k^a$. Note that if any $3$ of $x, z, xz, x/z$ are good, then the $4^{th}$ is good too $(1)$. We claim all numbers of the form $2^{c/2^b}$ where $c, b$ are non-negative integers, are good. We prove this by strong induction on $c$. The case $c = 0$ is trivial. For the case $c = 1$, we induct on $b$, with base case $b = 0$ trivially true, and the induction step following from $f(x) = \sqrt{f(x^2)+2}$. Returning to our induction on $c$, if the hypothesis is true till $c = l$, $l \ge 1$, the induction step follows by putting $x = 2^{l/2^b}, z = 2^{1/2^b}$ in $(1)$. Note that the function of $x$, $x^a + 1/x^a$ with domain $[1, \infty)$ has range $[2, \infty)$ and is strictly increasing for any positive real number $a$. Since $f$ is strictly increasing over $[1, \infty)$ and the good set of numbers of the form $2^{c/2^b}$ is a dense subset of $[1, \infty)$, all numbers in $[1, \infty)$ are good, thus all positive real $x$ are good.

Nice problem! The answer is $f(x) = x^r + \frac{1}{x^r}$ for $r \in \mathbb{R}$. These can be easily verified to work, so we show uniqueness. Let $P(x,y,z)$ denote the assertion. We have the following preliminary observations: Plugging in $P(1,1,1)$ yields $4f(1) = f(1)^3$, or $f(1) = 2$. Plugging in $P(x^2,1,1)$ yields $f(x^2) + 2 = f(x)^2$. Plugging in $P(x^2, \tfrac1{x^2}, 1)$ yields $$4 + f(x^2) + f(\tfrac1{x^2}) = 2f(x)f(\tfrac1x),$$and noting the LHS is equal to $f(x)^2 + f(\tfrac1x)^2$ gives $f(x) = f(\tfrac1x)$. We thus turn out attention to defining $f$ over the interval $[1,\infty)$. The final push is substituting $P(x^2,y^2,1)$, which yields $$f(x^2y^2) + f(x^2) + f(y^2) + 2 = f(xy)f(x)f(y).$$The LHS is equal to $f(xy)^2 + f(x)^2 + f(y)^2 - 4$, so the equation is a quadratic in $f(xy)$. Solving gives $$f(xy) = \frac{f(x)f(y) \pm \sqrt{(f(x)^2 - 4)(f(y)^2 - 4)}}{2}.$$I claim that for $x,y \geq 1$, the $\pm$ on the RHS is always a $+$. The claim is trivial for $x = 1$ and $y=1$, so assume $x,y > 1$. Without loss of generality $x \geq y$. Because $f(xy) > f(x)$, we must have $$f(x)f(y) - \sqrt{(f(x)^2 - 4)(f(y)^2 - 4)} > 2f(x) \implies f(x)(f(y)-2) > \sqrt{(f(x)^2-4)(f(y)^2-4)}.$$Squaring both sides and dividing by $f(y) - 2$ gives $$f(x)^2(f(y)-2) > (f(x)^2-4)(f(y)+2) \implies f(x)^2 < f(y) + 2,$$or $f(y) > f(x^2)$. This is absurd, as $y \leq x < x^2$, as desired. Finally, because $f(x) \geq 2$ for all $x \geq 1$, it makes sense to let $f(x) = g(x) + \frac{1}{g(x)}$ for a function $g : [1,\infty) \to [1,\infty)$. Substituting this into the quadratic formula yields $g(xy) = g(x)g(y)$. Because $f$ is increasing on $[1,\infty)$, so must $g$, and by Cauchy FE $g(x) = x^r$ as desired.

We claim that the answer is $f(x)=x^r+\frac{1}{x^r}$ for all $r\in\mathbb{R}_{> 0}$. Denote the assertion as $P(x,y,z)$. First, note that $P\left(x,x,\frac{1}{x}\right)$ yields $\left(f(1)^2-3\right)f(x)=f\left(\frac{1}{x}\right)$ and $x\to\frac{1}{x}$ gives $f(1)=2,f(x)=f\left(\frac{1}{x}\right)$ Now, $P\left(z,x,\frac{1}{x}\right)$ yields $2f(z)+2f(x)=f(1)f\left(\sqrt\frac{z}{x}\right)f\left(\sqrt{zx}\right)$. If we consider $z\to rz$, the change on the RHS is independent of $x$, so we can rewrite this as $f(a)f(b)-f(ar)f(br)$ for a given $r$ is constant if $ab$ is fixed. Setting $ab=1$, this means $f(x)^2-f(xr)f\left(\frac{x}{r}\right)=g(r)$ for some function $g$. Now, suppose $r>1$ and we fix $f(r)=k>2$. The function $h(m)=r^m+\frac{1}{r^m}$ for $m>0$ is surjective over $\mathbb{R}_{>2}$, so we can choose suitable $m$ such that $f(r)=r^m+\frac{1}{r^m}$. Now, as $g(r)=f(1)^2-f(r)^2$, $f(r)$ must determine all values of $f(r^z)$ for $z\in\mathbb{N}$ just by applying our equation over and over again. Thus, these values of $f$ are fixed. However, setting $f$ to be $r^m+\frac{1}{r^m}$ does satisfy the equation, and therefore we have $f(r^z)=r^{zm}+\frac{1}{r^{zm}}$. Now, consider $s>1$ and $f(s)=s^n+\frac{1}{s^n}$. Then, $f(s^z)=s^{zn}+\frac{1}{s^{zn}}$. Assume $m\neq n$ and WLOG $m<n$. Then, the function $x^n+\frac{1}{x^n}$ grows faster than $x^m+\frac{1}{x^m}$, so we can find suitable $z_1,z_2$ such that $s^{z_1}<r^{z_2}$ but $f(s^{z_1})>f(r^{z_2})$, which is a contradiction to the second condition. Thus, our assumption is wrong, and there must exist a constant $m$ such that $f(r)=r^m+\frac{1}{r^m}$ for all $r>1$, and we are done.

Solved with anser. The answer is $f(x) = x^r + \frac{1}{x^r}$ for some $r > 0$. Let $P(x, y, z)$ denote the equation. First, $P(1, 1, 1)$ gives $f(1) = 2$, and $P\left(x, x, \frac{1}{x}\right)$ gives $3f(x) + f\left(\frac{1}{x}\right) = 4f(x)$, so $f(x) = f\left(\frac{1}{x}\right)$. Also $P\left(x^2, 1, 1\right)$ gives $f(x^2) + 2 = f(x)^2$. Now we focus on $x > 1$, noting that $f(x) > f(1) = 2$ for $x > 1$. Claim: If $f(a) = a^k + \frac{1}{a^k}$ for some positive integer $k$, then $f(a^c) = a^{ck} + \frac{1}{a^{ck}}$ for all positive integers $c$. Proof: We use induction on $c$ with base case $c = 1$. If $c$ is even, then \[f(a^c) = f(a^{c/2})^2 - 2 = \left(a^{c/2} + \frac{1}{a^{c/2}}\right)^2 - 2 = a^{c} + \frac{1}{a^c}.\]If $c$ is odd, $P(a^{c-2}, a, a)$ gives $f(a^c) + f(a^{c-2}) + 2f(a) = f(a^{(c-1)/2})^2f(a) = (f(a^{c-1})+2)f(a)$, so \[f(a^c) = \left(a^{c-1} + \frac{1}{a^{c-1}}\right)\left(a + \frac{1}{a}\right) - \left(a^{c-2} + \frac{1}{a^{c-2}}\right) = a^{ck} + \frac{1}{a^{ck}}.\] Now since $f(x) > 2$, for any $x$ we can write $f(x) = x^k + \frac{1}{x^k}$ for some $k > 0$. Assume that this value of $k$ is not constant over all $x > 1$. Then there exists $x < y$ with $f(x) = x^k + \frac{1}{x^k}$ and $f(y) = y^\ell + \frac{1}{y^\ell}$. If $k < \ell$, then we can replace $y$ with $y^{1/t}$ for some very large $t$ such that $y^{1/t} < x$, and swap $x$ and $y$ (since $y^{1/t}$ and $y$ have the same $\ell$-value by the previous claim). So assume $k > \ell$. Claim: If $x, y > 1$, and $k > \ell$ are positive integers, then there exists positive integers $c$ and $d$ such that $x^c \leq y^d$ and $x^{ck} > y^{d\ell}$. Proof: The second equation rearranges to \[\left(\frac{x^c}{y^d}\right)^k > y^{d(\ell - k)}.\]No matter what $d$ is, we can choose $c$ such that $\frac{x^c}{y^d} \geq \frac{1}{x}$ (as if it's less than this, we can increase $c$ by $1$), so the LHS is at least $\frac{1}{x^k}$. Meanwhile, since $y > 1$ and $\ell - k < 0$, as $d$ gets large the RHS gets arbitrarily small. So we can pick $d$ extremely large to achieve this. But this contradicts the increasing condition -- we have $f(x^c) = x^{ck} + \frac{1}{x^{ck}}$ and $f(y^d) = y^{dk} + \frac{1}{y^{dk}}$ by the first claim. Since $x + \frac{1}{x}$ is increasing on $x > 1$, and $x^{ck} > y^{dk}$, this means $f(x^c) > f(y^d)$. But $1 \leq x^c < y^d$, so this is a contradiction. Remark: The motivation for the proof that $k$ must be constant is that if, for example, $f(3) = 3^6 + \frac{1}{3^6}$ and $f(4) = 4^5 + \frac{1}{4^5}$, then \[f(3^5) = 3^{30} + \frac{1}{3^{30}} > f(4^4) = 4^{20} + \frac{1}{4^{20}}\]even though $3^5 < 4^4$, which suggests that we can generally find a counterexample (to the second condition) of this form.

orl wrote: Let $\mathbb{R}^+$ be the set of all positive real numbers. Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ that satisfy the following conditions: - $f(xyz)+f(x)+f(y)+f(z)=f(\sqrt{xy})f(\sqrt{yz})f(\sqrt{zx})$ for all $x,y,z\in\mathbb{R}^+$; - $f(x)<f(y)$ for all $1\le x<y$. Proposed by Hojoo Lee, Korea Kinda easy. Let \(P(x,y,z)\) denote the assertion of the given functional equation. \(P(1,1,1)\) gives us that \(f(1)=2\) and \(P(x,1,1)\) gives us that \[f(\sqrt{x})^2=f(x)+2\]Let \(f(x)=g(x)+\frac{1}{g(x)}\) (we can do this since \(f(x)>f(1)=2\) for \(x\geq1\)). Then, the above relation re-writes as \[g(\sqrt{x})+\frac{1}{g(\sqrt{x})}=\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\]Now, \(P(x,y,1)\) combined with the fact above gives us that \[\left(\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\right)^2+\left(\sqrt{g(y)}+\frac{1}{\sqrt{g(y)}}\right)^2+\left(\sqrt{g(xy)}+\frac{1}{\sqrt{g(xy)}}\right)^2=\left(\sqrt{g(x)}+\frac{1}{\sqrt{g(x)}}\right)\left(\sqrt{g(y)}+\frac{1}{\sqrt{g(y)}}\right)\left(\sqrt{g(xy)}+\frac{1}{\sqrt{g(xy)}}\right)+4\]and so \(g(x)g(y)=g(xy)\), \(g(xy)g(x)=g(y)\) or \(g(x)g(y)g(xy)=1\) (see PFTB chap 1). All relations except \(g(x)g(y)= g(xy)\) give us that \(g(x)=1\) for all \(x\geq 1\) which is impossible. Therefore \(g\) is multiplicative and strictly monotone after \(1\), so \(g(x)=x^k\) for all \(x\geq 1\) where \(k\) is a fixed integer, so \(f(x)=x^k+\frac{1}{x^k}\) for all \(x\geq1\). Now let \(x_0<1\) be a real. Choose \(y_0, z_0\) arbitrarily large. Then \(P(x_0,y_0,z_0)\) gives us that \(f(x_0)=x_0^k+\frac{1}{x_0^k}\) since all inputs in the first condition except \(x_0\) are greater than \(1\) and you get the functional relation as an identity. In conclusion, the only solution is \(f(x)=x^k+\frac{1}{x^k}\) for any integer \(k\).

Let $P(x,y,z)$ be the given assertion. $P(1,1,1)$ gives $f(1)=2.$ $P(x,1/x,x)$ gives $f(x)=f(1/x).$ $P(1,x^2,1)$ gives $f(x^2)=f(x)^2-2.$ For Euler $e$ we have $f(e)\geq f(1)$ thus for constant $k,$ $f(e)=e^k+\frac{1}{e^k}.$ By induction for any $s,t \in \mathbb{Z^+}$ $$f(e^{s\frac{1}{2^t}})=e^{ks\frac{1}{2^t}}+\frac{1}{e^{ks\frac{1}{2^t}}}.$$Moreover $f$ is increasing provides $f(x)=x^k+\frac{1}{x^k},$ which fits.

We claim that the answer is $x^k+\frac{1}{x^k}$ which clearly follows the rules. Now to prove it is the only answer. $~$ Let $P(x,y,z)$ denote the assertion. Now, $P(x,x,x^{-1})$ gives $3f(x)+f(x^{-1})=4f(x)$ so $f(x^{-1})=f(x).$ $~$ Thus, while $f(x)\ge 2$ on $x\ge 1$ we also must have $f(x)\ge 2$ on $x\le 1$ so let $f(x)=\frac{1}{g(x)}+g(x).$ Then, $P(1,x^2,y^2)$ gives $f(x^2y^2)+f(x^2)+f(y^2)+2=f(x)f(y)f(xy).$ $P(1,1,x^2)$ gives $f(x^2)=f(x)^2-2$ so $f(xy)^2+f(x)^2+f(y)^2-4=f(x)f(y)f(xy).$ $~$ Now, when we substitute in $g(x)$ we get that $g(xy)=g(x)g(y).$ Combined with $g$ being increasing on interval $(1,\infty)$ and decreasing on interval $(0,1)$ we have $g(x)=x^k$ for some $k$ as desired.

The given FE can be written as \[f(abc)+f\left(\frac{bc}{a}\right)+f\left(\frac{ca}{b}\right)+f\left(\frac{ab}{c}\right)=f(a)f(b)f(c)\]Plugging $a=b=c=1$ gives $4f(1)=f(1)^3$ which forces, $f(1)=2$. $f(b^2c^2)+f(1)+f(b^2)+f(c^2)=f(bc)f(b)f(c)$ $4+f(b^2)+f(\frac1{b^2})=2f(b)f(c)$ $f(\frac{bc}a)+f(abc)+f(\frac{c}{ab})+f(\frac{b}{ca})=f(\frac1a)f(b)f(c)$ f(bc)+f(b/c)+f(c/b)=2f(b)f(c)$ f(b/c)+f(bc)+f(1/bc)=2f(b)f(1/c) f(c/b)-f(1/bc)=2f(b)(f(c)-f(1/c)) f(c)-f(1/c)=4(f(c)-f(1/c)) f(1/c)=f(c) f(b)+f(b)+f(1/b)+f(b)=4f(b) [just plug a=c=1] f(a)+f(1/a)+f(ab^2)+f(a/b^2)=f(a)f(b)^2=2f(a)+f(ab^2)+f(a/b^2) a=1 ==> f(b)^2 = 2 + f(b^2). f(x) >= 2 because strictly increasing. f(x) = g(x) + 1/g(x) f(xy)^2 + f(x)^2 + f(y)^2 - 4 = f(x)f(y)f(xy) Plug in g in terms of f, all terms cancel out and we're left with g(xy)=g(x)g(y). g is strictly increasing for x>1 and strictly decreasing for x<1, which means log (g):= cx, or g=x^c for some real number c>0. We are done.

The answer is $\boxed{x^k+\tfrac{1}{x^k}}$, which works. Let the given assertion be denoted as $P(x,y,z)$. Plug in $P(1,1,1)$ to get $P(1)=2$. Hence, for all $x \ge 1$, we have $f(x) \ge 2$. Moreover, $P(x,x,\tfrac{1}{x})$ yields that \[f(x)+f(x)+f(x)+f\left(\frac{1}{x} \right) = f(x)f(1)f(1) \implies f(x) = f\left(\frac{1}{x} \right).\] This means that $f(x) \ge 2$ on the interval $(0,1)$ as well, so we can set \[f(x) = g(x)+\frac{1}{g(x)}.\] Then, $P(1,x^2, y^2)$ and $P(1,1,x^2)$ give \[f(x^2y^2)+2+f(x^2)+f(y^2) = f(x)f(y)f(xy),\] and \[f(x^2)+4+f(x^2) = 2f(x)^2,\] respectively. The latter translates $f(x^2) = f(x)^2-2$, which when plugged into the former, yields \[(f(xy)^2-2)+(f(x)^2-2)+(f(y^2)-2) + 2 = f(x)f(y)f(xy).\] Now, refer back to the $g(x)$ substitution and denote $a = g(x)$, $b=g(y)$, and $c=g(xy)$: \[a^2 + \frac{1}{a^2} + b^2+\frac{1}{b^2}+c^2+\frac{1}{c^2} +2 = \left(a+\frac{1}{a} \right)\left(b+\frac{1}{b} \right)\left(c+\frac{1}{c} \right).\] Upon factoring, we see that either $abc=1$ or two of $a,b,c$ multiply to the other. The former is absurd, so we see that $g(xy)=g(x)g(y)$. Thus, $g(x) = x^k$, which implies our desired answer.

The solution is $f \equiv \boxed{x^k + \frac 1{x^k}\text{ for some }k>0.\text{ This is easy to verify.}}$ $(1, 1, 1) \implies f(1) = 2.$ $(x, 1, 1) \implies f(x) + 2 = f(\sqrt x)^2.$ Thus, $f \geq \sqrt{2}, $ but then $f\geq \sqrt{2 + \sqrt 2} \dots f \geq 2.$ Thus, define $g\colon[1, \infty)\to[1, \infty)$ so that $f = g + \frac1g$. $(x^2, y^2, 1) \implies f(xy)^2 + f(x)^2 + f(y)^2 - 4 = f(x)f(y)f(xy).$ By a quadratic in $f(xy),$ we have \[f(xy) = \frac{f(x)f(y) \pm \sqrt{(f(x)^2 - 4)(y^2 - 4)}}{2}\], so plugging in $f = g + \frac 1g$ gives \begin{align*} g(xy) + \frac 1{g(xy)} & = \frac{\left(g(x) + \frac 1{g(x)}\right)\left(g(y) + \frac 1{g(y)}\right) \pm \left(g(x) - \frac 1{g(x)}\right)\left(g(y) - \frac 1{g(y)}\right)}2 \\ & = g(x)g(y) + \frac1{g(x)g(y)}, \frac{g(x)}{g(y)} + \frac{g(y)}{g(x)} \end{align*}Thus, $g(xy) = g(x)g(y), \frac1{g(x)g(1)}, \frac{g(x)}{g(y)}, \frac{g(x)}{g(y)}$. However, only $g(xy) = g(x)g(y)$ satisfies satisfies the increasing condition. Thus, $g \equiv x^k$ since it is bounded below. Thus, $f \equiv x^k + \frac 1{x^k}$ for $x\geq 1.$ $(x, x, \frac 1x) \implies f(x) = f(\frac 1x).$ Thus, we are done.