<s>[INST] How prove this inequality $\sum\limits_{cyc}\frac{a^2}{b(a^2-ab+b^2)}\ge\frac{9}{a+b+c}$ let $a,b,c>0$, show that
$$\dfrac{a^2}{b(a^2-ab+b^2)}+\dfrac{b^2}{c(b^2-bc+c^2)}+\dfrac{c^2}{a(c^2-ca+a^2)}\ge\dfrac{9}{a+b+c}$$
My try: since this inequality is  homogeneous ,without loss of generality, we assume that
$$a+b+c=3$$
then 
$$\Longleftrightarrow  \sum_{cyc}\dfrac{a^2}{b(a^2-ab+b^2)}\ge 3$$
$$\Longleftrightarrow \sum_{cyc}\dfrac{a^2(a+b)}{b(a^3+b^3)}\ge 3$$
then I can't,Thank you
 [/INST]                      Proof without words ( partial answer / informal proof ) .
A formal proof has been given in another answer by another author.

Picture on the left: geometry of the conditions $a,b,c > 0$ and $a+b+c=3$ .
Picture in the middle: outside ($< 0$ : olive green) and inside ($\ge 0$ : white) of the function:
$$f(a,b,c) = \sum_{cyc}\frac{a^2}{b(a^2-ab+b^2)} - \frac{9}{a+b+c}$$
as seen in the plane of the red triangle in the picture on the left.
Since there are no green spots inside the triangle, the function is expected to be $\ge 0$ there.
Picture on the right: contour lines of $f(a,b,c)$ inside the triangle at levels $N = 1/2^k \;; \; k=0,\cdots,7$ . Contours are darker at lower level values.The minimum is expected to be $f(1,1,1)=0$ , which is at the center of the triangle.

</s>