Exponential-Monster Inequality

Let $a$ and $b$ be arbitrary positive numbers and $\alpha$ and $\beta$ two positive numbers such that $\alpha+\beta=1.$ Then the following inequality is always true due to the fact that the natural logarithm has a concave (or concave down) graph: $\ln(\alpha a +\beta b)> \alpha \ln a+\beta \ln b.$ Evaluating the function $y=e^x$ at the expressions that appear in the right side and the left side of the previous inequality, respectively, and using the fact that that function $y=e^x$ is increasing, it follows that, $\alpha a +\beta b >a^{\alpha}b^{\beta}.$ Dividing both sides by the positive number $\alpha a +\beta b,$ we get that $\frac{a^{\alpha}b^{\beta}}{\alpha a +\beta b }<1.$ Now, let us assume that $x<y<z$ and make $a=F(x, y),$ $b=F(y, z),$ $\alpha=\frac{y-x}{z-x},$ and $\beta=\frac{z-y}{z-x}.$ Then the following inequality follows $\large\frac{F(x, y)^{\frac{y-x}{z-x}}F(y, z)^{\frac{z-y}{z-x}}}{\frac{y-x}{z-x}F(x,y) +\frac{z-y}{z-x}F(y, z)}<1.$ Using direct computation you can see that the denominator of the fraction in the left side of the previous inequality reduces to $F(x, z).$ Then the previous inequality can be rewritten as $\large\frac{F(x, y)^{\frac{y-x}{z-x}}F(y, z)^{\frac{z-y}{z-x}}}{F(x, z)}<1.$ This inequality proves that $k$ exists and has to be smaller than or equal to $1.$ But we can prove that the value of $k$ is exactly $1$ by showing that when you fix a value of $x$ and a value of $z,$ such that $x<z,$ and you take the $y$ in between these two values, but making it tend to z, the limit of the left side of the previous inequality is 1. Then $k=1,$ and, therefore, the answer is $\boxed{1\leq k<2}.$