WLOG let $a=\min(a,b,c)$ Let $b=a+x, c=a+y$ and $x,y \ge 0$ Then the inequality is equivalent to \[2a(x^2-xy+y^2)+(x^3-3x^2y+2xy^2+y^3) \ge 0\] which is obvious

It's extremely nice. So we expand the inequality to $\sum_{cyc} (a^3+2a^2b-3b^2a)$. Draw the triangle even though it's really simple since this is through v_Enhance's SOS/Dumbassing article. Now there's nothing in the middle, which encourages us to look at each side individually. Since the corners have to go to two sides each, they parts on each side are going to have to add to $1$, whereupon $a-b$ divides the expression. So to clarify that bundle of thoughts, after WLOGing $b=1$ I would like to find an $x$ such that $xa^3+2a^2-3a+(1-x)\ge 0$. This would insta-solve our problem. Now divide out $a=1$ to get $(a-1)(x(a^2+a+1)+2a-1)$. If I want this nonnegative, I need another factor of $a-1$. Plug in $a=1$ so we want $x(1^2+1+1)+2(1)-1=0$ or $x=-\tfrac{1}{3}$. So we'll use $x=\tfrac{-1}{3}$ and factor. We get that this is $\tfrac{1}{3}(a-b)^2(4b-a)$. Unfortunately this is not always positive, but we have a sum of squares now! We desire to show that $\sum_{cyc} (a-b)^2(4b-a)\ge 0$. Now since this is cyclic we have two cases. WLOG $b$ is the median, so either $a\ge b\ge c$ or $c\ge b\ge a$. Now in the first case we have $S_b,S_a+S_b,S_b+S_c=4a-c,4a+3c-b, 3a+4b-c\ge 0$ so we are done by SOS criteria. So we only need to check $c\ge b\ge a$. As it turns out, this is really resistant to SOS criteria. But if we just return to the original inequality... it's trivial! Note that both $a(b-a)(2b-a),b(c-b)(2c-b)\ge 0$. So if we can show that $c(c-a)(c-2a)\ge 0$ we will be done. But note that if $c\le 4a$ then in the SOS form $S_a,S_b,S_c\ge 0$ so we win in this case. If $c\ge 4a$ then clearly the $c(c-a)(c-2a)\ge 0$ so we are done. SOS kills and is actually nice sometimes

NOTE: I had initially posted an incorrect solution. I realized my mistake and have adapted the proof accordingly, however, it is currently absurdly long . I'm working on shortening it.

A bashy proof: Let $(a-b)(b-c)(c-a)=P$. All sums are symmetric. The inequality is equivalent to \[\sum a^3 -\sum a^2b+5P\ge 0.\] If $P\ge 0$ we are done; otherwise, $P<0$, so $-P>0$ and it suffices to verify that \[\left( \sum a^3-\sum a^2b\right)^2\ge 25P^2.\] The left hand side is \[\left( 3(a^3+b^3+c^3)-(a^2+b^2+c^2)(a+b+c)\right)^2=\]\[=9(a^3+b^3+c^3)^2+(a^2+b^2+c^2)^2(a+b+c)^2-6(a^3+b^3+c^3)(a^2+b^2+c^2)(a+b+c).\] This is easy to expand using symmetric sums; it is \[\sum 2a^6-\sum 4a^5b-\sum 3a^4b^2+\sum a^4bc+\sum 5a^3b^3-\sum 2a^3b^2c+\sum a^2b^2c^2.\] The right hand side is \[25(a^2-2ab+b^2)(b^2-2bc+c^2)(c^2-2ac+a^2).\] This is also easy to expand by counting up the number of terms of a specific type that will emerge, e.g. counting all of the $a^4b^2$ terms. We find \[P^2=\sum a^4b^2-\sum a^4bc-\sum a^3b^3+\sum 2a^3b^2c-\sum a^2b^2c^2.\] Hence, we need to show that \[\sum a^6-\sum 2a^5b-\sum 14a^4b^2+\sum 13 a^4bc+\sum 15a^3b^3-\sum 26a^3b^2c+\sum 13a^2b^2c^2\ge 0,\] where we have subtracted and divided by $2$. In fact we can cancel off all $abc$ terms, so it suffices to show that \[\sum a^6-\sum 2a^5b-\sum 14a^4b^2+\sum 15a^3b^3\ge 0.\] This is constant in $abc$, so it suffices to check when $c=0, b=1$ and when $a=b=1$. In the first case, we need to show that \[f(x)=x^6-x^5-7x^4+15x^3-7x^2-x+1\ge 0.\] Note that $x^6f(\tfrac{1}{x})=f(x)$, so it suffices to prove for when $x\ge 1$. Let $x=e+1$, and we find \[f(e+1)=e^6+5e^5+3e^4-3e^3+e^2+3e+1\ge 0.\] When $a=b=1$, we need to show that \[f(x)=x^6-2x^5-x^4+4x^3-x^2-2x+1\ge 0.\] But \[f(x)=(x-1)^4(x+1)^2\ge 0,\] so we are done. $\square$

For #1 With condition $a,b,c>0$,we have \[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)=\frac{\sum{(59ac+3b^2+bc)(b+c-2a)^2}}{75(a+b+c)}+\frac{12\sum{(ab-2ac-b^2+bc+c^2)^2}}{25(a+b+c)}\ge{0}\]

szl6208 wrote: For #1 With condition $a,b,c>0$,we have \[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a)=\frac{\sum{(59ac+3b^2+bc)(b+c-2a)^2}}{75(a+b+c)}+\frac{12\sum{(ab-2ac-b^2+bc+c^2)^2}}{25(a+b+c)}\ge{0}\] Can you please elaborate on your solution? I'm assuming you expanded then factorized, but I have no idea how one can do this by hand

v_Enhance wrote: Let $a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\]Wenyu Cao Dear ,Best is this: Let $a,b,c$ be nonnegative real numbers. Prove that \[a \left( {a}^{2}-{b}^{2} \right) \left( a-2\,b \right) +b \left( {b}^ {2}-{c}^{2} \right) \left( b-2\,c \right) +c \left( {c}^{2}-{a}^{2} \right) \left( c-2\,a \right)\geq 0\] BQ

For #8 \[\sum{a(a^2-b^2)(a-2b)}=\frac{2abc}{a+b+c}\sum{(a-b)^2}+\frac{\sum{a(a^2-ab+ac-b^2+bc-c^2)^2}}{a+b+c}\ge{0}\]

v_Enhance wrote: Let $a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\]Wenyu Cao

v_Enhance wrote: Let $a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\]Wenyu Cao https://artofproblemsolving.com/community/c6h205316p1130156

WLOG $a\le b\le c$ or $a\ge b\ge c$. Expanding the above gives $\underset{\text{cyc}}{\sum}(a^3-3a^2b+2ab^2)=\underset{\text{cyc}}{\sum}a(a-b)^2+\underset{\text{cyc}}{\sum}(ab^2-a^2b)$. By realizing the right term is equal to $0$ when $a=b, b=c$, or $c=a$, we get that we want to show $\underset{\text{cyc}}{\sum}a(a-b)^2+(b-c)(c-a)(a-b)$. If $a\le b\le c$, both terms are positive as desired. If $a\ge b\ge c$, use Titu's Favorite Factoring Trick to rewrite our inequality as $\underset{\text{cyc}}{\sum}a(a-b)^2+(b-c)(c-a)(a-b)=\underset{\text{cyc}}{\sum}a(a-b)^2+\underset{\text{cyc}}{\sum}\frac{(a-b)^3}{3}=\underset{\text{cyc}}{\sum}(a-b)^2\left ( \frac{4a-b}{3}\right )$. If $a\ge 2b$, then if $b\ge 2c$ we are done. If $a\ge 2b$ and $b\le 2c$, then $a(a-b)(a-2b)\ge 0$, so it suffices to show that $c(a-c)(2a-c)\ge b(b-c)(2c-b)$. This is true because $c\ge 2c-b, a-c\ge b-c, 2a-c\ge b$. If $a\le 2b$, then $\frac{4a-b}{3}+2\cdot \frac{4c-a}{3}\ge 0$ and $\frac{4b-c}{3}+2\cdot \frac{4c-a}{3}\ge 0$, so by SOS we are done.

v_Enhance wrote: Let $a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\]Wenyu Cao $$(a^{2}+b^{2}+c^{2}-bc-ca-ab)LHS=a(a-b)^{4}+b(b-c)^{4}+c(c-a)^{4}\geq 0.$$

VMF-er wrote: v_Enhance wrote: Let $a,b,c$ be nonnegative real numbers. Prove that \[ a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0.\]Wenyu Cao $$(a^{2}+b^{2}+c^{2}-bc-ca-ab)LHS=a(a-b)^{4}+b(b-c)^{4}+c(c-a)^{4}\geq 0.$$ How can you find that ?

I think this was one of the first olympiad problems I solved. I had just read about BW on aops while mindlessly scrolling and lo and behold it actually worked WLOG let $a=\min(a,b,c)$, and write $b=a+x$, $c=a+y$ for $x,y \geq 0$. The inequality then becomes \begin{align*} ax(a+2x)+(a+x)(x-y)(x-a-2y)+y(a+y)(y-a)&\geq 0\\ 2ax^2-2axy+2ay^2+x^3-3x^2y+2xy^2+y^3&\geq 0\\ 2a((x-y)^2+xy)+x(x-2y)^2+y(x-y)^2&\geq 0, \end{align*}which is evident. $\blacksquare$

\[a(a - b)(a - 2b) + b(b - c)(b - 2c) + c(c - a)(c - 2a) \geq 0 \iff \sum{a^3}+2\sum{ab^2}\overset{?}{\geq} 3\sum{a^2b}\]Let $a/b=x,b/c=y,c/a=z$. Rewrite the inequality as the following with the condition $xyz=1$, \[\sum{x^2y}+2\sum{x}\overset{?}{\geq} 3\sum{yz}\]By Vasc, we have $\sum{x^2y}=\frac{\sum{x^2y}\sum{x}}{\sum{x}}=\frac{\sum{x^2y^2}+\sum{x^3y}}{\sum{x}}+1\geq \frac{\sum{x^2y^2}+\frac{(\sum{xy})^2}{3}}{\sum{x}}+1$. Let $x+y+z=3u,xy+yz+zx=3v^2,xyz=1$. \[\sum{x^2y}+2\sum{x}\overset{?}{\geq} 3\sum{yz}\iff \frac{9v^4-6u+3v^4}{3u}+1+6u\overset{?}{\geq} 9v^2\]\[9v^4-6u+3v^4+3u+18u^2\overset{?}{\geq} 27uv^2\iff 4v^4+6u^2\overset{?}{\geq} u+9uv^2\]We can see that $4v^4+6u^2=\underbrace{v^4+\dots+v^4}_{4}+\underbrace{u^2+\dots+u^2}_{6}\geq 10\sqrt[10]{v^{16}u^{12}}\geq 10uv\geq 9uv^2+u$ since $u\geq v\geq 1$ as desired.$\blacksquare$