augustin_p wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $$f(x)+f(x+y) \leq f(xy)+f(y).$$ Let $P(x,y)$ be the assertion $f(x)+f(x+y)\le f(xy)+f(y)$ Let $a=f(0)$ $P(x,0)$ $\implies$ $f(x)-a\le 0$ $\forall x$ $P(x,-x)$ $\implies$ $f(x)-f(-x)\le f(-x^2)-a\le 0$ $P(-x,x)$ $\implies$ $f(-x)-f(x)\le f(-x^2)-a\le 0$ And so $f(-x^2)=a$ and $f(x)=f(-x)$ $\forall x$ And so $\boxed{f(x)=a\quad\forall x}$, which indeed fits whatever is $a\in\mathbb R$

As usual let $P(x,y):=f(x)+f(x+y)\le f(xy)+f(y)$ $P(x,0)$ yields $2f(x)\le2f(0)\Longrightarrow f(x)\le f(0)$ $P(-x^2,0)$ yields $2f(-x^2)\le2f(0)\Longrightarrow f(-x^2)\le f(0)$ (!!) $P\left(\frac{x^4}{4},0\right)$ yields $f\left(\frac{x^4}{4}\right)\le f(0)$ (!!) Furthermore $P(x,-x)$ yields $f(x)+f(0)\le f(-x^2)+f(-x)$ and $P(-x,x)$ yields $f(-x)+f(0)\le f(-x^2)+f(x)$ Now, notice that from $P(x,-x)+P(-x,x)$ we obtain $$f(x)+f(0)+f(-x)+f(0)\le f(-x^2)+f(-x)+f(-x^2)+f(x)\Longrightarrow f(x)+f(-x)+2f(0)\le 2f(-x^2)+f(-x)+f(x)\Longleftrightarrow f(0)\le f(-x^2)$$However, notice that from (!!) we have that $f(0)\le f(-x^2)\le f(0)$, which forces $f(-x^2)=f(0)\Longrightarrow f(\text{negative})=f(0)$ Furthermore notice that from $P\left(-\frac{x^2}{2},-\frac{x^2}{2}\right)$ we obtain $f(-x^2)=f(0)\le f\left(\frac{x^4}{4}\right)$, however from (!!) we have that $f(0)\le f\left(\frac{x^4}{4}\right)\le f(0)$ which forces $f\left(\frac{x^4}{4}\right)=f(0)\Longleftrightarrow f(\text{positive})=f(0)$ thus $f(x)=f(0), \forall x\in\mathbb{R}$ In conclusion $\boxed{f(x)=c, \forall x\in\mathbb{R}\text{ and any constant }c\in\mathbb{R}}$ $\blacksquare$.

augustin_p wrote: Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfy the following condition for any real numbers $x{}$ and $y$ $$f(x)+f(x+y) \leq f(xy)+f(y).$$ $\color{blue}\boxed{\textbf{Answer: }f\equiv c}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(x)+f(x+y) \leq f(xy)+f(y)...(\alpha)$$In $(\alpha) y\to 0:$ $$\Rightarrow 2f(x)\leq 2f(0)$$$$\Rightarrow f(x)\leq f(0)...(\beta)$$In $(\alpha) y\to -x:$ $$\Rightarrow f(x)+f(0)\leq f(-x^2)+f(-x)...(i)$$By $(\beta):$ $$\Rightarrow f(x)+f(-x^2)\leq f(x)+f(0) \leq f(-x^2)+f(-x)$$$$\Rightarrow f(x)\leq f(-x)...(\theta)$$In $(\theta) x\to -x:$ $$\Rightarrow f(-x)\leq f(x)$$$$\Rightarrow f(x)=f(-x)...(\omega)$$Replacing $(\omega)$ in $(i):$ $$\Rightarrow f(0)\leq f(-x^2)$$By $(\beta):$ $$\Rightarrow f(-x^2)=f(0)$$By $(\theta):$ $$\Rightarrow f(x)=f(0),\forall x\in\mathbb{R}$$$$\Rightarrow \boxed{f(x)\equiv c}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

f(x)=c with easy substitution

ismayilzadei1387 wrote: f(x)=c with easy substitution Excellent contribution

tadpoleloop wrote: ismayilzadei1387 wrote: f(x)=c with easy substitution Excellent contribution Hahah i didint want to post same substitution.