Find all functions $f\colon\mathbb{R}\to\mathbb{R}$ such that $(x-y)\bigl(f(x)+f(y)\bigr)\leqslant f\bigl(x^2-y^2\bigr)$ for all $x,y\in\mathbb{R}$.
Problem
Source: BxMO 2023, Problem 1
Tags: BxMO, algebra
06.05.2023 18:23
14.05.2023 12:40
Let the given assumption be denoted by $P(x,y)$ By comparing $P(x,x)$ and $P(x,0)$ we infer that $f(0)=0$ Then by $P(x,-x)$ we know that $2x(f(x)+f(-x)) \le 0$ so by $x \rightarrow -x$ we obtain $2x(f(x)+f(-x)) \geq 0$ and thus $2x(f(x)+f(-x))=0$ which implies $f$ is odd. Next , swapping $x,y$ we obtain $(y-x)(f(x)+f(y)) \le f(y^2-x^2)=-f(x^2-y^2) \implies f(x^2-y^2) \le (x-y)(f(x)+f(y)) \implies f(x^2-y^2)=(x-y)(f(x)+f(y))$ So by $y \rightarrow -y$ we get $(x+y)(f(x)-f(y))=f(x^2-y^2)=(x-y)(f(x)+f(y)) \implies \frac {f(x)}{x}=\frac{f(y)}{y}$ for all $x,y \in \mathbb{R}$ So $\frac{f(x)}{x}$ is constant and hence $f(x)=kx$ for some constant $k$ which clearly satisfies the condition.
12.06.2023 23:20
yeppp just plug necessary valuables and find $-f(x)=f(-x)$ so $P(x,y)$ - $P(x,-y)$=$xf(y)=yf(x)$ plug $y=1$ and $f(x)=ax$ $a\in \mathbb{R}$
13.06.2023 00:15
Let $P(x,y):=(x-y)(f(x)+f(y))\le f(x^2-y^2)$ $P(x,x)$ yields $f(0)\ge0$ $P(1,0)$ yields $f(0)\le0$ thus $f(0)=0$ $P(x,-x)$ yields $2x(f(x)+f(-x))\le0$ furthermore by $x\rightarrow -x$ we get that $2x(f(x)+f(-x))\ge0$, thus $2x(f(x)+f(-x))=0\Longrightarrow -f(x)=f(-x)$ therefore $f$ is odd, also since $f$ is odd we must have equality. $P(x,-y)$ yields $(x+y)(f(x)-f(y))=f(x^2-y^2)$ furthermore from $P(x,y)$ we obtain $(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y)$, after expanding we are left with $xf(y)=yf(x)$ and let $Q(x,y)$ denote $xf(y)=yf(x)$ $Q(x,1)$ yields $f(x)=f(1)x\Longrightarrow f(x)=cx$ for some constant $c\in\mathbb{R}$ So, to sum up $\boxed{f(x)=cx, \forall x \in\mathbb{R}\text{ and for some constant } c}$ $\blacksquare$
25.07.2023 14:31
x--->y f(0)>=0 x->1 , y->0 f(0)=<0 both of them gives f(0)=0 y-> -x 2x(f(x)+f(-x))=<0 x->-y -2y(f(y)+f(-y))=<0 in last inequality y->x 2x(f(x)+f(-x))>=0 both of last two inequalities we can find function is odd Quote: other x=0 in problem y-> -y
function is odd for that we can solve this problem for R+-->R P(x^0,5.y^0,5) gives f(x-y)>=f(x)-f(y) P(y,x) (f(x)+f(y))(x-y) -1=<-1f(x^2-y^2) P(x,y) showes openent from that (f(x)+f(y))(x-y)=f(x^2-y^2) f(x^2-y^2)-f(x^2)+f(y^2)= f(y)x-yf(x) >=0 f(y)x>=yf(x) x->1 f(x) >=cx y->1 f(x)=< cx both of them gives f(x)=cx checking : c(x^2-y^2)=< c(x^2-y^2)
17.01.2024 12:07
Let $P(x,y)$ be the assertion. $P(1,0)$ yields: $$f(1) + f(0) \leq f(1) \implies f(0) \leq 0$$$P(0,0)$ gives $f(0) \geq 0 \implies f(0) = 0$. $P(x,-x)$ gives: $$2x(f(x) + f(-x)) \leq 0 \implies -2x(f(x) + f(-x)) \geq 0$$$P(-x, x)$ gives $-2x(f(x) + f(-x)) \leq 0 \implies f(x) + f(-x) = 0$. $P(y,x)$ yields: $$(y - x)(f(y) + f(x)) \leq f(y^2 - x^2) = -f(x^2 - y^2)$$$P(x,y)$ gives $(x-y)\bigl(f(x)+f(y)\bigr)\leqslant f\bigl(x^2-y^2\bigr)$. From these two, we have equality. $P(x, -1)$ and $P(x,1)$ give: $$(x + 1)(f(x) - f(1)) = (x - 1)(f(x) + f(1)) \implies f(x) = xf(1) \implies f(x) = cx$$Substituting it back into the original equation, we see that it works hence $f(x) = cx$.
19.09.2024 18:29
The answer is $f(x) = cx$ for some fixed constant $c\in \mathbb{R}$ for all $x\in \mathbb{R}$. It’s easy to see that these functions satisfy the given equation. We now show these are the only solutions. Let $P(x,y)$ be the assertion that $f(x^2-y^2) \geq [f(x)+f(y)](x-y)$ . We start off by making some preliminary observations. Note that $P(0,0)$ gives $f(0)\ge 0$ and $P(1,0)$ gives $f(1) \ge f(1) +f(0)$ so $f(0)\le 0$. Combining these two inequalities implies that we must have $f(0)=0$. Next, $P(x,-x)$ and $P(-x,x)$ for $x>0$ imply, \[0=f(0) \ge 2x(f(x)+f(-x))\]and \[0 = f(0) \ge -2x(f(x)+f(-x)) \implies 0 \le 2x(f(x)+f(-x))\]Thus, $2x(f(x)+f(-x))=0$ which implies that $f(x)=-f(-x)$ for all $x>0$ (and in turn for all $x\ne 0$). Now, we prove the following key result. Claim : For all functions $f$ which satisfy the given condition, in fact we must also have \[f(x^2-y^2)=[f(x)+f(y)](x-y)\]for all real numbers $x$ and $y$. Proof : Say there exists $x_0,y_0 \in \mathbb{R}$ such that $f(x_0^2-y_0^2) \ne [f(x_0)+f(y_0)](x_0-y_0)$. Now, $P(x_0,y_0)$ and $P(y_0,x_0)$ give \[f(x_0^2-y_0^2)> [f(x_0)+f(y_0)](x_0-y_0)\](since as per our assumption equality does not hold), and \[f(y_0^2+x_0^2) \ge [f(y_0)+f(x_0)](y_0-x_0)\]Adding the above two inequalities gives, \[f(x_0^2-y_0^2)+f(y_0^2-x_0^2) > (f(x_0)+f(y_0))(x_0-y_0+y_0-x_0)=0\]Thus, $f(x_0^2-y_0^2)+f(y_0^2-x_0^2) > 0$ which is a clear contradiction to our previous observation that $f(x)=-f(-x)$ for all $x\in \mathbb{R}$. Thus, such a pair of real numbers $(x_0,y_0)$ cannot exist, proving the desired claim. Now, we simply need to find all solutions to the functional equation, \[f(x^2-y^2)=[f(x)+f(y)](x-y)\]for all real numbers $x$ and $y$. For this, note that $P(x,-y)$ and $P(-x,y)$ imply \[f(x^2-y^2)=[f(x)+f(-y)](x+y)\]\[f(x^2-y^2)=[f(-x)+f(y)](-x-y)\]Summing these two equations then gives, \begin{align*} 2f(x^2-y^2) &= (x+y)(f(x)+f(-y)-f(-x)-f(y))\\ 2f(x^2-y^2) &= (x+y)(f(x)-f(y)+f(x)-f(y))\\ f(x^2-y^2) &= [f(x)-f(y)](x+y) \end{align*}But this then implies, \[[f(x)-f(y)](x+y) = f(x^2-y^2) = [f(x)+f(y)](x-y)\]which upon expanding and simplifying reduces to, \[\frac{f(x)}{x}=\frac{f(y)}{y}\]for all $x,y\ne 0$. Considering $y=1$ implies that $f(x)=xf(1)=cx$ for all $x\ne 0$ for some fixed constant $c \in \mathbb{R}$. Thus, all solutions to the given functional inequality are indeed of the claimed forms.
09.11.2024 00:00
Denote $P(x,y)$ the assertion of the F.I. By $P(0,0)$ we get $f(0) \ge 0$, from $P(x,-x)$ we get $2x(f(x)+f(-x)) \le f(0)$ for all reals $x$. $P(x,0)$ gives $f(x^2) \ge x(f(x)+f(0))$, so $x=1$ here gives $0 \ge f(0)$ so $f(0)=0$ which means that $2x(f(x)+f(-x)) \le 0$ for all reals $x$, so swap $x \to -x$ to get $2x(f(x)+f(-x)) \ge 0 \ge 2x(f(x)+f(-x))$ therefore $f(x)+f(-x)=0$ for all reals $x$ (yes even $0$), which means that $f$ is odd. Now do $P(y,x)$ and multiply by $-1$ to get that the F.E. turns out to be an Equality so $(x-y)(f(x)+f(y))=f(x^2-y^2)$ for all $x,y \in \mathbb R$, therefore we get $f(x^2)=xf(x)$ for all reals $x$, but also if you swap $y \to -y$ here we get $xf(x)+xf(y)-yf(x)-yf(y)=xf(x)-xf(y)+yf(x)-yf(y)$ which after simplying becomes $xf(y)=yf(x)$ so set $y=1$ here to get $f(x)=xf(1)$ for all reals $x$. Therefore the only family of solutions are $f(x)=cx$ for a constant $c$ and all reals $x$, thus we are done .
09.11.2024 05:11
The answer is $f(x)\equiv 0$ and $f(x)=x$. If $f(x)\equiv c$, then $2c(x-y)\leq c,\forall x,y\in \mathbb{R}$, then $c=0$. If $f(x)$ is non-constant. Let $P(x,y)$ be the assertion in the hypothesis. $P(x,x)$ gives $f(0)\geq 0$ and $P(1,0)$ gives $f(0)\leq 0$, so $f(0)=0$. $P(x,-x)$ gives $2x(f(x)+f(-x))\leq f(x^2-x^2)=0,\forall x\in \mathbb{R}$. Then $$x(f(x)+f(-x))\leq 0,\forall x\in \mathbb{R}.$$In this equation, put $x\to -x$, we have $-x(f(x)+f(-x))\leq 0,\forall x\in \mathbb{R}$. Then $f(x)+f(-x)=0,\forall x\in \mathbb{R}$, or $f$ is odd. $P(y,x)$ gives $f(y^2-x^2)\geq (y-x)(f(x)+f(y)),\forall x,y\in \mathbb{R}$, so $-f(x^2-y^2)\geq -(x-y)(f(x)+f(y)),\forall x,y\in \mathbb{R}$. Then $f(x^-y^2)\leq (x-y)(f(x)+f(y)),\forall x,y\in \mathbb{R}$. Compare this equation with problem, we get $$(x-y)(f(x)+f(y))=f(x^2-y^2),\forall x,y\in \mathbb{R}.$$In this equation, put $y\to -y$, we get $(x+y)(f(x)-f(y))=f(x^2-y^2),\forall x,y\in \mathbb{R}$. Then $$(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y)),\forall x,y\in \mathbb{R}.$$Thus, $xf(y)=yf(x),\forall x,y\in \mathbb{R}$. Then there exists a constant real number $d$ such that $f(x)=dx,\forall x\in \mathbb{R}$. Easily to check that $f(x)=dx$ is satisfy.