john111111 wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ which satisfy $yf(x)+f(y) \geq f(xy)$ Let $P(x,y)$ be the assertion $yf(x)+f(y)\ge f(xy)$ Let $c=\frac{f(-1)}2$ $P(-1,x)$ $\implies$ $f(x)-f(-x)\ge -2cx$ $P(-1,-x)$ $\implies$ $-2cx\ge f(x)-f(-x)$ So $f(-x)=f(x)+2cx$ $P(x,y)$ $\implies$ $yf(x)\ge f(xy)-f(y)$ $P(x,-y)$ $\implies$ $f(y)+2cy-f(xy)-2cxy \ge yf(x)$ So $f(y)+2cy-f(xy)-2cxy\ge f(xy)-f(y)$ which is $f(y)+cy\ge f(xy)+cxy$ and so $f(x)+cx$ is constant over $\mathbb R^*$ Plugging this back in original equation, we get $\boxed{f(x)=a(1-x)}$ $\forall x$ (even $x=0$) which indeed is a solution, whatever is $a\in\mathbb R$

Let $P(x;y)$ be the assertion of $yf(x)+f(y)\ge f(xy)$, By substituting $P(1;1)$ and $P(1;-1)$ we get that $ f(1)=0$. As well as by $P(y;x)$ $f(xy\leq x f(y)+f(x))$. So $P(x;xy)$ leads to $f(x^2y)\leq xy f(x)+f(xy)\leq xy f(x)+x f(y)+f(x)$. By $P(-x;\frac{-1} {x})$ one may easily find out that $f(x) + x f(\frac{1} {x})\leq 0$.So by putting $P(\frac{1} {y}; y)$ we have that $f(x) + x f(\frac{1} {x})\geq 0$. Hence $f(x) + x f(\frac{1} {x})=0$. By $P(x/y ; y)$ we obtain that $y f(\frac{x} {y})+f(y)\geq f(x)$ and by $P(x;xy)$ we get that $xy f(\frac{1} {y})+f(xy)\geq f(x)$. We have that $f(x) + x f(\frac{1} {x})=0$, which pays the way for $f(xy)\geq nf(x)+xf(y)$ and of course we got $f(xy)\leq x f(y)+f(x)$. Therefore $f(xy)=x f(y)+f(x)$ for all $x,y R$. Which leads to $f(xy)=y f(x)+f(y)=x f(y)+f(x)$. By putting $x=2$ to this equation we can get that $f(y)=f(2)(y-1)$ where $a\in\mathbb R$. It is easy to verify that this is indeed a solution.

john111111 wrote: Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ which satisfy $yf(x)+f(y) \geq f(xy)$ Now can you find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ which satisfy $y^2f(x)+f(y) \geq f(xy)$ ?

Here is another solution: Let $P(x,y)$ be the assertion $yf(x)+f(y)\ge f(xy)$ and let $f(0)=c$: $P(0,x) \Longrightarrow f(x)\ge (1-x)c$ $(1)$ So, $yf(x)+f(y)\ge f(xy) \ge (1-xy)c \Longrightarrow yf(x)+f(y)\ge (1-xy)c$ $(*)$ Plugging $y=x$ in $(*)$, we get: $$(x+1)f(x) \ge (1-x)(1+x)c$$Therefore, if $x<-1$, we have that: $$f(x) \le (1-x)c$$This combined with $(1)$ gives that $f(x)=(1-x)c$ for all $x<-1$. Plugging in $(*)$ $y<-1$ and in light of the previous result, we have (note that $y<0$): $yf(x)+(1-y)c\ge (1-xy)c \Longleftrightarrow yf(x)\ge (y-xy)c \Longleftrightarrow$ $$f(x)\le (1-x)c$$So, $f(x)=(1-x)c$ for all $x\in\mathbb R$, where $c$ is a constant with $c\in\mathbb R$.

Here is another solution: Take $y\neq 0$. $P(x,y) + P(x,-y) \implies f(y) + f(-y) \geq f(xy) + f(-xy) \implies f(y) + f(-y) \geq f(z) + f(-z) \geq f(y) + f(-y)$ for all non-zero $y,z$. Hence $f(y) + f(-y) = c$ for all non-zero $y$. So for all $x,y \neq 0$ equality holds in $P(x,y) + P(x,-y) \implies yf(x) + f(y) = f(xy)$ for all $x,y \neq 0$. $P(1,y) + P(-1, y) \implies 2f(y) = c(1-y) \implies f(y) = a(y-1)$. Now $P(0,y) \implies (y-1)f(0) + a(y-1) \geq 0$ for all $y \neq 0$. Hence $a+f(0) = 0 \implies f(0) = -a$. Hence $f(x) = a(x-1)$ for all $x$, and checking this works.