Prove inequality using a given inequality Question: It is given that $$a^r(a-b)(a-c)+b^r(b-c)(b-a)+c^r(c-a)(c-b)\geq 0$$ for positive integers $a,b,c,r$. We are supposed to show that $$\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}+\frac{a+b+c}{a^2b^2c^2}\geq \frac{b^2+c^2}{a^3b^2c^2}+\frac{c^2+a^2}{a^2b^3c^2}+\frac{a^2+b^2}{a^2b^2c^3}.$$
Attempt: I tried to multiply $a^5b^5c^5$ throughout in an effort to "reverse engineer" and hopefully, see the answer. What I got was $$b^5c^5+a^5c^5+a^5b^5+a^3b^3c^3(a+b+c)\geq a^2b^3c^3(b^2+c^2)+a^3b^2c^3(c^2+a^2)+a^3b^3c^2(a^2+b^2).$$ Then I tried to let $x=bc$, $y=ac$ and $z=ab$ so that I have $x^5+y^5+z^5+\dots\geq\dots$ and hopefully this would provide an expression that is identical to letting $r=5$. Unfortunately, I didn't succeed.
Any hints or suggestions would be much appreciated!

Alternative (brute force/buffalo way) solution: Your inequality is upon full expansion (using symmetric sum notation) $$\frac{\frac{\sum_{\text{sym}} a^5 b^5}2+\frac{\sum_{\text{sym}}a^4 b^3 c^3}2-\sum_{\text{sym}}a^5 b^3 c^2}{a^5b^5c^5}\ge 0.$$
Since $a,b,c>0$, you just need to prove $$\frac{\sum_{\text{sym}} a^5 b^5}2+\frac{\sum_{\text{sym}}a^4 b^3 c^3}2-\sum_{\text{sym}}a^5 b^3 c^2\ge 0$$ or $$\sum_{\text{sym}} a^5 b^5+\sum_{\text{sym}}a^4 b^3 c^3- 2 \sum_{\text{sym}}a^5 b^3 c^2\ge 0.$$
Based on how the exercise is posed, they seem to want you to find some smart substitution to turn this into Schur's inequality. A more brute force approach is this: Assume without loss of generality that $a\le b$ and $a\le c$. We may then write $b=a+x, c=a+y$ for some $x,y\ge 0$. Then the  last inequality becomes upon full expansion $$8 a^8 x^2\color{red}-8 a^8 x y+8 a^8 y^2+8 a^7 x^3+20 a^7 x^2 y+20 a^7 x y^2+8 a^7 y^3+2 a^6 x^4+24 a^6 x^3 y+118 a^6 x^2 y^2+24 a^6 x y^3+2 a^6 y^4+6 a^5 x^4 y+140 a^5 x^3 y^2+140 a^5 x^2 y^3+6 a^5 x y^4+66 a^4 x^4 y^2+178 a^4 x^3 y^3+66 a^4 x^2 y^4+12 a^3 x^5 y^2+92 a^3 x^4 y^3+92 a^3 x^3 y^4+12 a^3 x^2 y^5+18 a^2 x^5 y^3+50 a^2 x^4 y^4+18 a^2 x^3 y^5+10 a x^5 y^4+10 a x^4 y^5+2 x^5 y^5\ge 0$$ which is true since $$8a^8 x^2+8a^8 y^2 \ge 16 a^8 xy$$ by AM-GM.
