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Let $\alpha,\beta,\gamma$ be three real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that $$\alpha\,f(yz)+\beta\,f(zx)+\gamma\,f(xy)\geq f(x+y+z)$$ for all $x,y,z\in\mathbb{R}$.

Remarks.

  • If $\alpha=\beta=\gamma=0$, then any function $f:\mathbb{R}\to\mathbb{R}_{\leq0}$ is a solution.

  • If $\alpha+\beta+\gamma=1$, then any constant function $f$ satisfies the functional inequality.

  • If $\alpha+\beta+\gamma>1$, then any nonnegative constant function $f$ satisfies the functional inequality.

  • If $\alpha+\beta+\gamma<1$, then any nonpositive constant function $f$ satisfies the functional inequality.

Special Case. (The idea was borrowed from this answer, which is a particular case with $\alpha=\beta=\gamma=\dfrac13$.)

Suppose that $\alpha+\beta+\gamma=1$. Without loss of generality, $\alpha\leq \beta\leq \gamma$. Plugging in $y:=0$ and $z:=0$, we get $$f(x)\leq f(0)$$ for all $x\in\mathbb{R}$. Note that $\gamma>0$. Plugging in $y:=-x$ and $z:=0$, we have $$f(0)\geq f(-x^2)\geq f(0)$$ for all $x\in\mathbb{R}$. This means $f(x)=f(0)$ for all $x\leq 0$. Now, by letting $y:=x$ and $z:=-2x$, we have $$f(0)\geq f(+x^2)\geq f(0)$$ for all $x\in\mathbb{R}$. This means $f(x)=f(0)$ for all $x\geq 0$. Therefore, $f(x)=f(0)$ for every $x\in\mathbb{R}$. Hence, if $\alpha,\beta,\gamma\in\mathbb{R}$ are such that $\alpha+\beta+\gamma=1$, then the only solutions $f$ are constant functions.

What are solutions for other parameters $(\alpha,\beta,\gamma)$? Can we hope to find all solutions $f$ for an arbitrary triple $(\alpha,\beta,\gamma)$? If this is too difficult, can we at least hope for a characterization of all solutions $f$ if $\alpha=\beta=\gamma$?

Postscript. If we have an equality instead, that is, $$\alpha\,f(yz)+\beta\,f(zx)+\gamma\,f(xy)=f(x+y+z)$$ for all $x,y,z\in\mathbb{R}$, then by plugging $y,z:=0$, we get $f(x)=(\alpha+\beta+\gamma)\,f(0)$. In particular, when $x:=0$, we have $$(\alpha+\beta+\gamma-1)\,f(0)=0\,.$$ Thus, $\alpha+\beta+\gamma=1$ or $f(0)=0$.

If $\alpha+\beta+\gamma=1$, then using the same argument as the proof of the "Special Case" above, we can see that $f(x)=f(0)$ for all $x\in\mathbb{R}$. That is, $f$ is a constant function.

If $\alpha+\beta+\gamma\neq 1$, then $f(0)=0$. Using the same argument as the proof of the "Special Case" above, we can see that $f(x)=0$ for all $x\in\mathbb{R}$. That is, $f$ is the zero function.

Batominovski
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    I did $\alpha=\beta=\gamma=1$ and $f(t) = \frac{1}{1+t^2} ; . ; ;$ Seems promising – Will Jagy Aug 21 '20 at 23:54
  • hmmm it kind of appears that the highest degree term in the resulting numerator is $-x^4 y^4 z^4$ while denominator has degree 14 as $(x^2 + y^2 + z^2)x^4 y^4 z^4,$ if that holds up the difference can be made negative at large $x=y=z.$ That should be easy enough to check... – Will Jagy Aug 22 '20 at 01:49
  • yeah, with $x=y=z=t$ the difference has positive denominator but numerator $2 + 27 t^2 - t^4$ – Will Jagy Aug 22 '20 at 01:56

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