prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$ prove $$\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$$ where $a,b,c>0$ and $a\ge b\ge c$
My try it seemed qite simple but i couldnt  apply the rearrangement inequality directly.  so  i tried manipulating the inequality.
the inequality can be written as $$a^2(b-c)+b^2(c-a)+c^2(a-b)\ge 0$$ .It seemed like 'schurs' inequality could be uused but i couldn't procceed.Also i tried using the  weighted a-m g-m method. Could anyone give me a hint (i want to solve the problem myself).
source: Excursion in mathematics(Modak)

Your first step is wrong:
We need to prove that:
$$\sum_{cyc}(a^3b^2-a^3bc)\geq0$$ and it indeed gives a proof:
$$\sum_{cyc}(2a^3b^2-2a^3bc)\geq0$$ or
$$\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)\geq\sum_{cyc}(a^3c^2-a^3b^2)$$ or
$$\sum_{cyc}a^3(b-c)^2\geq(ab+ac+bc)(a-b)(b-c)(c-a),$$ which is obvious.
We can get $$\sum_{cyc}(a^3c^2-a^3b^2)=(ab+ac+bc)(a-b)(b-c)(c-a)$$ by the following way.
For $a=b$, $a=c$ and  $b=c$ we obtain identity, which says that $$\sum_{cyc}(a^3c^2-a^3b^2)=P(a,b,c)(a-b)(b-c)(c-a),$$ where $P$ is a cyclic homogeneous polynomial of second degree.
Id est $$P(a,b,c)=\sum_{cyc}(ka^2+mab).$$
Now, $k=0$ because, otherwise there is a problem on $\infty$.
Also, let $c=0$.
We obtain: $$b^3a^2-a^3b^2=mab(a-b)b(-a)$$ or
$$a^2b^2(b-a)=ma^2b^2(b-a),$$ which gives $m=1$.
Factoring of some Schur's polynomials:
$$\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^3b-a^3c)=(a+b+c)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^4b-a^4c)=(a^2+b^2+c^2+ab+ac+bc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^3b^2-a^3c^2)=(ab+ac+bc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^5b-a^5c)=$$
$$=(a^3+b^3+c^3+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+abc)(a-b)(a-c)(b-c),$$
$$\sum_{cyc}(a^4b^2-a^4c^2)=(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc)(a-b)(a-c)(b-c),...$$
