dame dame

dame dame

Almost same solution as @above but whatever.... Let $P(a,b,c)$ denote the given assertion. $P(1,1,1)$ gives that $(f(1)-1)^2 \le 0$ which forces $f(1) = 1$ $P(a,b,1)$ gives that $f(a) + f(b) \ge 1 + f(ab)$ $P(1,x,x)$ gives that $f(x) + f(x^2) - f(x)^2 \ge 1$ Using the fact that $2f(x) -1 \le f(x^2)$ from the thing 2 lines above, we get that $3f(x) - f(x)^2 \ge 2$, which rearranges to $(f(x)-1)(f(x)-2) \le 0$, which means that $f(x) \in (1,2)$ for all $x \in \mathbb{N}$ So now, suppose the maximum value of $f(n)$ is $m$ and let $f(N) = m$ Then, $P(N,N,N)$ gives that $1 \le f(N^2)(2-f(N)) \le m(2-f(N)) \implies f(N) \le 2 - \frac{1}{m} \implies m + \frac{1}{m} \le 2$, which forces $m = 1$. But since we had that $f(x) \ge 1$ for all $x$, this means that $f(x) = 1$ for all $x$, which indeed is a solution

L567 wrote: Almost same solution as @above but whatever.... Let $P(a,b,c)$ denote the given assertion. $P(1,1,1)$ gives that $(f(1)-1)^2 \le 0$ which forces $f(1) = 1$ $P(a,b,1)$ gives that $f(a) + f(b) \ge 1 + f(ab)$ $P(1,x,x)$ gives that $f(x) + f(x^2) - f(x)^2 \ge 1$ Using the fact that $2f(x) -1 \le f(x^2)$ from the thing 2 lines above, we get that $3f(x) - f(x)^2 \ge 2$, which rearranges to $(f(x)-1)(f(x)-2) \le 0$, which means that $f(x) \in (1,2)$ for all $x \in \mathbb{N}$ So now, suppose the maximum value of $f(n)$ is $m$ and let $f(N) = m$ Then, $P(N,N,N)$ gives that $1 \le f(N^2)(2-f(N)) \le m(2-f(N)) \implies f(N) \le 2 - \frac{1}{m} \implies m + \frac{1}{m} \le 2$, which forces $m = 1$. But since we had that $f(x) \ge 1$ for all $x$, this means that $f(x) = 1$ for all $x$, which indeed is a solution MathForesterCycle1 wrote: Mr.C wrote: Find all functions $f : \mathbb{N} \rightarrow \mathbb{R}$ such that for all triples $a,b,c$ of positive integers the following holds : $$f(ac)+f(bc)-f(c)f(ab) \ge 1$$ Proposed by Mojtaba Zare Darn it! We see that the only constant function is $f \equiv 1$ so let us say $f$ is not constant. We see that $P(1, 1, 1) \implies f(1) = 1$. We also see that $P(a, a, a) \implies f(a^2)(2 - f(a)) \geq 1 - (\bigstar)$ and $P(a, a, 1) \implies 2f(a) - f(a^2) \geq 1 - (\spadesuit)$, putting $(\bigstar)$ and $(\spadesuit)$ together we get that $(2f(a)-1)(2-f(a)) \geq f(a^2)(2-f(a)) \geq 1$ implies that $f(a) \in [1, \frac{3}{2} ]$ for all $a \in \mathbb{Z}^+$. Then $P(a, a, a) \implies f(a^2)(2-f(a)) \geq 1$ and so $f(a^2) \geq \frac{1}{2-f(a)}$ but $f(a^2) \leq \frac{3}{2} \implies 2-f(a) \geq \frac{2}{3} \implies f(a) \leq \frac{4}{3}$. Now $f(a^2) \leq \frac{4}{3} \implies 2 - f(a) \geq \frac{3}{4} \implies f(a) \leq \frac{5}{4}$. So in fact, if there is some $a \in \mathbb{Z}^+$ such that $f(a) > 1$, let $a_i$ be such a positive integer such that $f(a_i)$ is maximized then there exists $n \in \mathbb{Z}^+$ such that $\frac{n}{n-1} >f(a_i) \geq \frac{n+1}{n} \implies 2 - f(a_i) \leq \frac{n-1}{n}$, $f(a_i^2) \geq \frac{1}{2-f(a)} \geq \frac{n}{n-1}$ which contradicts maximality of $f(a_i)$ and so $f \equiv 1$ as desired. Please check this proof as I am new to FEs There is no reason that "The biggest value" exist. it can be fixed with induction though

Similar to above.

dame dame

Mr.C wrote: There is no reason that "The biggest value" exist. it can be fixed with induction though Alternatively, replacing $\max$ with $\sup$ works fine.

$\overline{\underline{\textit{SOLUTION}}}$ Let $P(a,b,c) \Longrightarrow f(ac) + f(bc) - f(c)f(ab) \geq 1 $ $\boxed{\textbf{Claim 1.}} \ 1 \leq f(x) \leq 2 \ \text{and} \ f(1)=1$ $P(1,1,1) \Longrightarrow \left(f(1)-1 \right)^2 \leq 0$ which cause $f(1)=1$. $P(a,a,1) \Longrightarrow 2f(a) \geq f(a^2) + 1$ $P(a,1,a) \Longrightarrow f(a) + f(a^2) \geq f(a)^2 +1 $ By combining the 2 inequalities, we obtain \[ (f(a)-2)(f(a)-1) \leq 0 \]Which cause $1 \leq f(a) \leq 2$. $\rule{20 cm}{1 pt}$ $\boxed{\textbf{Claim 2.}}$ There doesn't exist $x \in \mathbb{N}$ such that $f(x) > 1$. Define the function $g(x) = f(x)-1 \Longrightarrow 0 \leq g(x) \leq 1$. From $f(a) + f(a^2) \geq f(a)^2 + 1$ we can obtain that, \[ g(a^2) \geq g(a) + g(a)^2\]Now, if there exist such $x$, then $g(x) >0$. From the above inequality, it is obvious that the value of $g(x^{2^k})$ for some arbitrarily large $k \in \mathbb{N}$ will also be arbitrarily large, which leads to a contradiction. So, we get that $f \equiv 1$ which is obviously a solution.

Let $P(a,b,c)$ be the inequality $$ f(ac)+f(bc)-f(c)f(ab)\geq 1 \quad \forall \, a,b,c\in \mathbb N. $$ Then, $P(1,1,1)$ gives $$ 2f(1)-f(1)^2\geq 1 \Longrightarrow f(1)=1 $$Now, $P(a,b,1)$ gives $$ f(a)+f(b)-f(ab)\geq 1 \quad \forall \, a,b\in \mathbb N. $$And $P(a,1,b)$ gives $$ f(ab)+f(b)-f(b)f(a)\geq 1 \quad \forall \, a,b\in \mathbb N. $$Adding up the two inequalities gives $$ f(a)+2f(b)-f(a)f(b)\geq 2 \Longrightarrow (f(a)-2)\cdot (f(b)-1)\leq 0 \quad \forall \, a,b\in \mathbb N. $$From this last inequality we get $1\leq f(x)\leq 2$. Now, we will prove $f\equiv 1$. Let $k\leq 2$ be an upper bound of $f$. Setting $f(c)=f(ab)=k$ in the original equation, one gets $$ f(ac)+f(bc)\geq k^2+1 $$But since $1\leq f(x)\leq 2$, then $k^2+1\leq 4 \Longrightarrow k\leq \sqrt{3} \Longrightarrow 1\leq f(x)\leq \sqrt{3}$. Similarly, setting $f(c)=f(ab)=k$ in the original equation, one gets $$ f(ac)+f(bc)=k^2+1 \Longrightarrow k^2+1\leq 2\sqrt{3} \Longrightarrow k\leq \sqrt{2\sqrt{3}-1} $$Hence $1\leq f(x)\leq \sqrt{2\sqrt{3}-1}$. Let $g(x)=\sqrt{2x-1}$. Then, by induction, it holds $$ k\leq \lim_{n\rightarrow \infty}{g^n(2)} $$In fact, the value of such limit is $1$.

Now, we know $$ \max f =k\leq 1 $$But indeed $1\leq f(x)\leq 2$. Therefore, $\boxed{f\equiv 1}$, which indeed fits.

Let show the main inequality as $P(a,b,c)$ , $$ P(1,1,1) \implies f(1)=1 $$$$ P(1,1,x) \implies f(x)\geq 1 $$$$ P(x,x,x) \implies f(x^2)\cdot (2-f(x))\geq 1 \implies 1\leq f(x)\leq 2 $$Now, by induction on $n$, we gonna show that $f(x)\in [1,\frac{n+1}{n} ] \quad \forall x\in \mathbb N $ which clearly implies $f\equiv 1.$ Let assume $$ f(x)\leq \frac{n+1}{n} $$is true. Than $$ P(x,x,x)\implies 1\leq (2-f(x))\cdot f(x^2)\leq (2-f(x))\cdot \frac{n+1}{n} \implies 2-\frac{n}{n+1} \geq f(x) $$$$ \implies 1\leq f(x)\leq \frac{n+2}{n+1} $$completes the induction. As a result $f\equiv 1$ is the only solution.

$P(x,y,z)\implies f(xz)+f(yz)-f(z)f(xy) \ge 1$ $P(1,1,1)\implies f(1)=1$ $P(x,x,1)+P(1,x,x)\implies f(x)\in [1,2]$ $\exists \beta : \max (f(x))= \beta$ $P(x,x,x)\implies \beta +\beta ^{-1} \leq 2$ $\therefore \beta =1\implies f(x)\equiv 1 \; \text{(works)}$

GuvercinciHoca wrote: Let show the main inequality as $P(a,b,c)$ , $$ P(1,1,1) \implies f(1)=1 $$$$ P(1,1,x) \implies f(x)\geq 1 $$$$ P(x,x,x) \implies f(x^2)\cdot (2-f(x))\geq 1 \implies 1\leq f(x)\leq 2 $$Now, by induction on $n$, we gonna show that $f(x)\in [1,\frac{n+1}{n} ] \quad \forall x\in \mathbb N $ which clearly implies $f\equiv 1.$ Let assume $$ f(x)\leq \frac{n+1}{n} $$is true. Than $$ P(x,x,x)\implies 1\leq (2-f(x))\cdot f(x^2)\leq (2-f(x))\cdot \frac{n+1}{n} \implies 2-\frac{n}{n+1} \geq f(x) $$$$ \implies 1\leq f(x)\leq \frac{n+2}{n+1} $$completes the induction. As a result $f\equiv 1$ is the only solution. It looks nice

Here is a difference solution ! Let show the main inequality as $P(a,b,c)$ : Calim 1 : $f(1) = 1$ Proof : $P(1,1,1):f(1)^2+1-2f(1) \leq 0 \Rightarrow f(1)=1$ Calim 2 : $1 \leq f(1) \leq 2$ Proof : $P(1,1,c) : f(x) \geq 1$ and $P(x,x,x) : 2f(x^2)-f(x)f(x^2) \geq 1$ if $f(x) > 2$ then $f(x^2) < 0$ but $f(x^2) \geq 1$ $$P(x,x^2,1) : f(x)+f(x^2)-f(x^3) \ge 1 (*)$$$$P(x,x^2,x) : f(x^2) \ge (f(x)-1)f(x^3)+1 (**)$$using $* , **$ we get : $f(x)+(f(x)-1)f(x^3)+1+f(x^3) \ge 1 \rightarrow f(x)(f(x^3)+1) \ge 2f(x^3)$ $$P(x,x,x^2) : 2f(x^3) \ge 1+f(x^2)f(x^2)$$$$P(x,x,x):f(x^2)\ge \frac{1}{2-f(x)}$$now : $$2f(x) \ge 2f(x^3) \ge 1+f(x^2)^2 \ge 1+\frac{1}{(2-f(x))^2}$$$$\rightarrow (2f(x)-1)(2-f(x))^2-1 \ge 0$$$$\rightarrow 2f(x)^3-9f(x)^2+10f(x)-5 \ge 0$$now let $g(y) = 2y^3-9y^2+10y-5$, the roots of $g'(x)=6y^2-18y+10$ are $2.26,0.73$ so $g$ is decreasing in the interval $[0.73,2.62]$ and $g(1)=0$ so $\forall 1<x<2 : g(x) < 0 $ it means $f(x)=1$

Solution from Twitch Solves ISL: The answer is $f \equiv 1$ only, which works. Conversely, let $P(a,b,c)$ be the statement. From $P(1,1,1)$ we get $0 \ge \left( f(1)-1 \right)^2 \implies \boxed{f(1)=1}$ From $P(1,1,c)$ we get $f(c) \ge 1$. Then, note that $P(x,x,x)$ gives the statement \begin{align*} f(x^2)+f(x^2)-f(x)f(x^2) &\ge 1 \\ \implies \left( 2-f(x) \right) f(x^2) &\ge 1 \\ \implies 2-f(x) &\ge \frac{1}{f(x^2)} \\ \implies f(x) &\le 2 - \frac{1}{f(x^2)} \end{align*}since $f$ only takes positive values. In particular, this implies that we have $f(n) \le 2$ for all $n \ge 1$; we have $f(n) \le \frac 32$ for all $n \ge 1$ (by using the previous statement to get $f(n^2) \le 2$); we have $f(n) \le \frac 43$ for all $n \ge 1$ (again by previous); we have $f(n) \le \frac 54$ for all $n \ge 1$ (again by previous); and so on. It follows that $f(n) = 1$. Remark: Note that the problem statement is only ever used when $a=b$.

$P(1,1,1) \Longrightarrow 0 \ge (f(1)-1)^2 \Longrightarrow f(1) = 1$ $P(1,1,c) \Longrightarrow f(c) \ge 1$ $P(a,a,a) \Longrightarrow f(a^2) + f(a^2) - f(a)f(a^2) \ge 1 \Longrightarrow (2-f(a))f(a^2) \ge 1 \Longrightarrow f(a) \le 2 - \frac{1}{f(a^2)} $ From this we get: $f(n) \le 2$ $f(n) \le \frac{3}{2}$ (By using $f(n^2) \le 2$ in $f(a) \le 2- \frac{1}{f(a^2)}$) $f(n) \le \frac{4}{3}$ and so on thus $f(n) = 1$

Let $P(a,b,c)$ be the assertion. From $P(0,0,0)$ we get $0\geq \left(f(0)-1\right)^{2},$ thus $f(0)=1.$ From $P(a,b,0)$ we get $2-f(ab)\geq1$ and $1\geq f(x)$$(*)$ for all $x \in \mathbb{N}.$ From $P(1,1,1)$ we have $0\geq \left(f(1)-1\right)^{2},$ thus $f(1)=1.$ From $P(1,1,x)$ we have $f(x)\geq1.(**)$ Lastly, from $(*)$ and $(**)$ we get $1\geq f(x)\geq 1,$ thus the only solution is $\boxed{f\equiv1}$ which clearly fits.