A geometric inequality?

Since the three numbers $x,$ $y,$ and $z$ are positive numbers, then we can always construct a triangle like the one above. We divide the given triangle into four triangles represented in the figure. Let $K$ denote the area of the big triangle $ABC$ . Then we have that $\sin{A}=\frac{2K}{(x+y)(x+z)},$ $\sin{B}=\frac{2K}{(y+x)(y+z)},$ and $\sin{C}=\frac{2K}{(z+x)(z+y)} (*).$ Clearly, the sum of the areas of the three small triangles that share vertices with the big triangle is less than the area of the big triangle. Hence, we have that $\frac{1}{2}x^2 \sin{A}+\frac{1}{2}y^2 \sin{B}+\frac{1}{2}z^2 \sin{C}<K.$ Now, using (*) and dividing both sides of the inequality by $K,$ we obtain that $\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}<1.$ But 1 is the smallest possible number for which the inequality is always true for any positive values of $x, y,$ and $z$ . Indeed, if you fix $y$ and $z$ and you make $x \rightarrow 0,$ then the expression in the left side of the inequality tends to 1.

The previous solution can be interesting due to its connection with geometry, but there is an even simpler solution here. It is based on the following reasoning. Since the numbers $x, y$ and $z$ are positive then $0< 2xyz$ Adding the quantity $x^2y+x^2z+y^2x+y^2z+z^2x+z^2y$ to both sides of the inequality, and rearranging them conveniently, we get $x^2(y+z)+y^2(x+z)+z^2(x+y)<(x+y)(x+z)(y+z).$ Dividing both sides by $(x+y)(x+z)(y+z),$ $\frac{x^2}{(x+y)(x+z)}+\frac{y^2}{(y+x)(y+z)}+\frac{z^2}{(z+x)(z+y)}<1.$ And we can also prove that 1 is the smallest possible number for which the inequality is always true as we did it above.