How Zeb intended this problem to die

MellowMelon wrote: Prove that for positive real numbers $ x,y,z$, \[ x^3(y^2 + z^2)^2 + y^3(z^2 + x^2)^2 + z^3(x^2 + y^2)^2 \geq xyz[xy(x + y)^2 + yz(y + z)^2 + zx(z + x)^2].\] It's easy SOS.

MellowMelon wrote: Prove that for positive real numbers $ x,y,z$, \[ x^3(y^2 + z^2)^2 + y^3(z^2 + x^2)^2 + z^3(x^2 + y^2)^2 \geq xyz[xy(x + y)^2 + yz(y + z)^2 + zx(z + x)^2].\] Zarathustra (Zeb) Brady. Let $ x=1/a,y=1/b,z=1/c$, the inequality becomes $ \sum \frac{(b^2+c^2)^2}{a^3b^4c^4} \ge\frac{1}{abc}\sum \frac{(b+c)^2}{b^3c^3},$ or equivalently, $ \sum a(b^2+c^2)^2 \ge \sum a^3(b+c)^2.$ (1) We have $ \sum a(b^2+c^2)^2 =\sum a(b^4+c^4)+2\sum ab^2c^2 =\sum a^4(b+c) +\sum a^2bc(b+c),$ so the last inequality can be written as $ \sum [a^4(b+c)+a^2bc(b+c)-a^3(b+c)^2] \ge 0,$ that is $ \sum a^2(b+c)(a-b)(a-c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS-RHS=\sum a(b-c)^2(b+c-a)^2.$

can_hang2007 wrote: $ \sum a^2(b + c)(a - b)(a - c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS - RHS = \sum a(b - c)^2(b + c - a)^2.$ I think the following generalization holds for any $ k\ge 0$ and nonnegative $ a,b,c$: $ \sum (b + c)(a - b)(a - c)(a - kb)(a - kc)\ge 0,$ with equality for $ b = c = a$, for $ b = c = 0$, for $ b = c = \frac a{k}$ (or any cyclic permutation). We have also a nice identity.

can_hang2007 wrote: $ \sum a^2(b + c)(a - b)(a - c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS - RHS = \sum a(b - c)^2(b + c - a)^2.$ This is an interesting stronger inequality $ \sum a^2(b + c)(a - b)(a - c) \ge \frac {4(a-b)^2(b-c)^2(c-a)^2}{a+b+c}.$

Vasc wrote: can_hang2007 wrote: $ \sum a^2(b + c)(a - b)(a - c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS - RHS = \sum a(b - c)^2(b + c - a)^2.$ I think the following generalization holds for any $ k\ge 0$ and nonnegative $ a,b,c$: $ \sum (b + c)(a - b)(a - c)(a - kb)(a - kc)\ge 0,$ with equality for $ b = c = a$, for $ b = c = 0$, for $ b = c = \frac a{k}$ (or any cyclic permutation). We have also a nice identity. The inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc=0$ or $ (a-b)(b-c)(c-a)=0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done.

can_hang2007 wrote: Vasc wrote: can_hang2007 wrote: $ \sum a^2(b + c)(a - b)(a - c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS - RHS = \sum a(b - c)^2(b + c - a)^2.$ I think the following generalization holds for any $ k\ge 0$ and nonnegative $ a,b,c$: $ \sum (b + c)(a - b)(a - c)(a - kb)(a - kc)\ge 0,$ with equality for $ b = c = a$, for $ b = c = 0$, for $ b = c = \frac a{k}$ (or any cyclic permutation). We have also a nice identity. The inequality is in degree 5th, therefore if we transform it into $ pqr$ form, the degree of $ r$ will be $ 1$, from which follows that we only need to prove it in the case $ abc = 0$ or $ (a - b)(b - c)(c - a) = 0$ (according from an theorem which I have posted already on the forum). But in both cases, the inequality is trivial, so we are done. Yes, Can_hang. We have also $ \sum (b + c)(a - b)(a - c)(a - kb)(a - kc)=\sum a(b-c)^2(b+c-(k+1)a)^2$.

can anyone help me to prove that $ \sum a^r(b+c)(a-b)(a-c) \ge 0$ where a,b,c,r are positive real. i cant prove it using vornicu schur because i cant understand vornicu schur

ishfaq420haque wrote: can anyone help me to prove that $ \sum a^r(b + c)(a - b)(a - c) \ge 0$ where a,b,c,r are positive real. i cant prove it using vornicu schur because i cant understand vornicu schur The following reasoning is strongest for the Vornicu-Schur's method. Let $ a\geq b\geq c.$ Hence, $ \sum_{cyc}a^r(b + c)(a - b)(a - c)\geq$ $ \geq a^r(b + c)(a - b)(a - c) + b^r(a + c)(b - a)(b - c) =$ $ = (a - b)(a^r(b + c)(a - c) - b^r(a + c)(b - c)) =$ $ = (a - b)((ab - c^2)(a^r - b^r) + c(a - b)(a^r + b^r))\geq0.$

MellowMelon wrote: Prove that for positive real numbers $ x,y,z$, \[ x^3(y^2 + z^2)^2 + y^3(z^2 + x^2)^2 + z^3(x^2 + y^2)^2 \geq xyz[xy(x + y)^2 + yz(y + z)^2 + zx(z + x)^2].\] Zarathustra (Zeb) Brady. Here is my solution \[ {{x}^{3}}{{\left( {{y}^{2}}+{{z}^{2}} \right)}^{2}}+{{y}^{3}}{{\left( {{z}^{2}}+{{x}^{2}} \right)}^{2}}+{{z}^{3}}\left( {{x}^{2}}+{{y}^{2}} \right)\ge xyz\left[ xy{{\left( x+y \right)}^{2}}+yz{{\left( y+z \right)}^{2}}+zx{{\left( z+x \right)}^{2}} \right]\] \[ \Leftrightarrow x[{{(x{{y}^{2}}+x{{z}^{2}})}^{2}}-{{({{y}^{2}}z+y{{z}^{2}})}^{2}}]+y[{{(y{{z}^{2}}+y{{x}^{2}})}^{2}}-{{({{x}^{2}}z+x{{z}^{2}})}^{2}}]+z[{{(z{{x}^{2}}+z{{y}^{2}})}^{2}}-{{({{x}^{2}}y+x{{y}^{2}})}^{2}}]\ge 0\] \[ \Leftrightarrow x(x{{y}^{2}}+x{{z}^{2}}+{{y}^{2}}z+y{{z}^{2}})[{{y}^{2}}(x-z)+{{z}^{2}}(x-y)]+y(y{{x}^{2}}+y{{z}^{2}}+{{x}^{2}}z+x{{z}^{2}})[{{x}^{2}}(y-z)+{{z}^{2}}(y-x)]+z(z{{x}^{2}}+z{{y}^{2}}+{{x}^{2}}y+x{{y}^{2}})[{{x}^{2}}(z-y)+{{y}^{2}}(z-x)]\ge 0\] Or $ x^2 (y-z)(x^2 y^2+yz^2+x^2 yz+xyz^2-x^2 z^2-z^2 y^2-x^2 yz-xy^2 z)+y^2 (z-x)(x^2 z^2+z^2 y^2+x^2 yz+xy^2 z-x^2 y^2-x^2 z^2-xy^2 z-xyz^2 )+z^2 (x-y)(x^2 y^2+z^2 x^2+xy^2 z+xyz^2-x^2 y^2-y^2 z^2-x^2 yz-xyz^2 )\geq 0$ $ \Leftrightarrow x^3 (y-z)^2 (xy+xz-yz)+y^3 (z-x)^2 (xy+yz-zx)+z^3 (x-y)^2 (yz+zx-xy)\geq 0$ And we have S.O.S method: $ S=S_x (y-z)^2+S_y (z-x)^2+S_z (x-y)^2\geq 0$ with \[ {{S}_{x}}={{x}^{3}}\left( xy+zx-yz \right),{{S}_{y}}={{y}^{3}}\left( xy+yz-zx \right),{{S}_{z}}={{z}^{3}}(yz+zx-xy)\] Assume \[ x\ge y\ge z\] \[ \Rightarrow {{S}_{y}}\ge 0\] We have \[ {{S}_{x}}+{{S}_{y}}={{x}^{4}}y+{{x}^{3}}z\left( x-y \right)+x{{y}^{4}}+{{y}^{3}}z\left( y-x \right)\ge z\left( x-y \right)\left( {{x}^{3}}-{{y}^{3}} \right)=z{{\left( x-y \right)}^{2}}\left( {{x}^{2}}+xy+{{y}^{2}} \right)\ge 0,\] \[ {{S}_{y}}+{{S}_{z}}={{y}^{4}}z+x{{y}^{3}}\left( y-z \right)+y{{z}^{4}}+x{{z}^{3}}\left( z-y \right)\ge x\left( y-z \right)\left( {{y}^{3}}-{{z}^{3}} \right)=x{{\left( y-z \right)}^{2}}\left( {{y}^{2}}+yz+{{z}^{2}} \right)\ge 0\] Thus, \[ S=\left( {{S}_{x}}+{{S}_{y}} \right){{\left( y-z \right)}^{2}}+\left( {{S}_{y}}+{{S}_{z}} \right){{\left( x-y \right)}^{2}}+2{{S}_{y}}\left( x-y \right)\left( y-z \right)\ge 0\] The end.

Vasc wrote: can_hang2007 wrote: $ \sum a^2(b + c)(a - b)(a - c) \ge 0.$ This can be proved easily using Vornicu Schur. Especially, we have a very nice identity which implies (1), it is $ LHS - RHS = \sum a(b - c)^2(b + c - a)^2.$ This is an interesting stronger inequality $ \sum a^2(b + c)(a - b)(a - c) \ge \frac {4(a - b)^2(b - c)^2(c - a)^2}{a + b + c}.$ Assume : $ c=min\{a;b;c\}$ With : $ a=x+c \ ; \ b=y+c \ ; \ x,y \geq 0$ It is equivalent to : $ 6c^4x^2-6c^4xy+6c^4y^2+14c^3x^3-9c^3x^2y-9c^3xy^2+14c^3y^3+10c^2x^4+c^2x^3y-18c^2x^2y^2+c^2xy^3+10c^2y^4+2cx^5+5cx^4y-7cx^3y^2-7cx^2y^3+5cxy^4+2cy^5+x^5y-4x^4y^2+6x^3y^3-4x^2y^4+xy^5 \geq 0$ Which is true

Is the Muirhead inequality allowed here? well even if its not its very easy to make it with A-G (4,3,0)+(3,2,2)$ \geq$ (4,2,1)+(3,2,2) and this is obvious? have I maybe miss understood the task?(it seems too easy)

Standard mistake, I thought also that Muirhead kills it. It's not 4,3,0 + 3,2,2 $ \ge$ 4,2,1 + 3,2,2 It's 4,3,0 + 3,2,2 $ \ge$ 4,2,1 + 3,3,1 and that isn't possible with Muirhead. But maybe, when multiplying with x+y+z (brilliant idea if it's true, see post #2) it will reduce to something solvable with Schur/Muirhead...

you are right my mistake sorry and thanks for showing it to me this x+y+z is very very nice idea but I didn't manage to solve it after doing that so could someone provide that solution?

Hm... I also tried multiplying, but we are left with (5,2,1)+2(4,3,1) $ \le$ (5,3,0)+(4,4,0)+(3,3,2) Murihead can't do it, Schur applies only with (a,0,0)... Maybe it's something else.

Does anyone has a link where is explained Vornicu Schur method I've never heard of it. Muirhead solves it if you square the ineq you have (8,6,0)+(7,7,0)+2(7,4,3)+3(6,4,4)+(8,3,3)+4(5,5,4)+2(7,5,2)+2(6,6,2)+2(6,5,3) >= (8,4,2)+3(6,6,2)+2(6,5,3)+3(5,5,4)+(8,3,3)+4(6,5,3)+2(7,5,2)+2(7,4,3) is equal as (8,6,0)+(7,7,0)+(5,5,4)>=(8,4,2)+(6,6,2)+(6,4,4) LHS>=(8,4,2)+2(6,6,2)>=RHS first time is muihead + A-G second just muirhead if I made another error please correct me cause this is done manually by hand so I would be suprised if its correct.

Why, and how did you use AM-GM on (x,y,z) ??? For Vornicu-Schur, search http://www.mathlinks.ro/portal.php?t=162684

thanks for the link and Am-GM is used to prove that (7,7,0)+(5,5,4)>=2(6,6,2) $ x^7y^7+x^5y^5z^4\geq 2 x^6y^6z^2$ $ y^7z^7+y^5z^5x^4\geq 2 y^6z^6x^2$ $ z^7x^7+z^5x^5y^4\geq 2 z^6x^6y^2$ and summing this 3 inequalities you get this ineq up here. And the question why I don't understand (is there a way to do it without it(I couldn't find it)?)

The inequality is equivallent to \[ x^{3}(y - z)^{2}(y + z) + y^{3}(z - x)^{2}(z + x) + z^{3}(x - y)^{2}(x + y)\geq{x^{2}yz(y - z)^{2} + xy^{2}z(z - x)^{2} + xyz^{2}(x - y)^{2}}.\] . Dividing both sides by $ xyz$ we get: \[ \frac {(xy - zx)^{2}(xy + xz - yz)}{yz} + \frac {(yz - xy)^{2}(yz + xy - zx)}{zx} + \frac {(xz - yz)^{2}(zx + zy - xy)}{xy}\geq{0}.\] . Substitute $ xy = a, yz = b, zx = c$. The inequality reduces to $ \frac {(a - b)^{2}(a + b - c)}{c} + \frac {(b - c)^{2}(b + c - a)}{a} + \frac {(c - a)^{2}(c + a - b)}{b}\geq{0}$. Rewrite the inequality as $ \frac {a(a - b)^{2}}{c} + \frac {b(a - b)^{2}}{c} + \frac {b(b - c)^{2}}{a} + \frac {c(b - c)^{2}}{a} + \frac {a(c - a)^{2}}{b} + \frac {c(c - a)^{2}}{b}\geq{(a - b)^{2} + (b - c)^{2} + (c - a)^{2}}.$ Substitute $ |a - b| = m$, $ |b - c| = n$, $ |c - a| = k$. The last inequality is equivallent to $ (m^{2}\frac {a}{c} + n^{2}\frac {c}{a}) + (m^{2}\frac {b}{c} + k^{2}\frac {c}{b}) + (n^{2}\frac {b}{a} + k^{2}\frac {a}{b})\geq{m^{2} + n^{2} + k^{2}}.$ Using AM-GM, $ LHS\geq{2mn + 2mk + 2kn}$, but easy to prove that $ m,n,k$ are the lengths of a triangle(maybe degenerate),or all of them is $ 0$. (To prove this, wlog assume $ m$ is maximum between $ (x,y,z)$. Than, by the well known inequality $ |x| + |y|\geq{|x + y|}$, $ k + n = |b - c| + |c - a|\geq{|(b - c) + (c - a)| = |a - b| = m}$, q.e.d.) So, \[ LHS\geq{2mn + 2nk + 2km} = (m(n + k - m) + n(k + m - n) + k(m + n - k)) + (m^{2} + n^{2} + k^{2})\geq{m^{2} + n^{2} + k^{2}},\] Q.E.D.

We've to show $\sum x^3(y^4+z^4)+2\sum x^3y^2z^2\geq xyz(\sum xy(x^2+y^2)+2\sum(xy)^2)$. Now suppose $xyz=1$ and lagrange function, $f(x,y,z,\lambda)=\sum xy(\frac {1}{x^4}+\frac{1}{y^4})+2\sum x-(\sum\frac {x^2+y^2}{z}+2\sum\frac{1}{x^2})$. After differentiating both side respectively WRT $x,y,z$ we obtain, $-\frac{3y+3z}{x^4}+\frac{y^3+z^3}{y^3z^3}+2-4x^3-2x(\frac{1}{y}-\frac{1}{z})+\frac{y^2+z^2}{x^2}=\lambda yz$ and similarly others. So, we've like, $-\frac {3a}{x^3}+bx+2x-4x^4-2x^2c+\frac {d}{x}+\frac {2}{x^2}+x$ is constant for $x,y,z$ where $a,b,c,d$ are same for $x,yz$. Now notable part is, again after derivative WRT $x,y,z$ we obtain, $-\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\frac {d'}{x}+\frac {2}{x^2}+x$ is constant for $x,y,z$ where $a',b',c',d'$ are same for $x,yz$ , hence we've like, $b''x+2x-4x^4-2x^2c''+\frac {d''}{x}+\frac {2}{x^2}+e'x$ is constant for $x,y,z$ where $b'',c'',d'',e'$ are same for $x,yz$. Again doing same with $-\frac {9a}{x^3}+b'x+2x-4x^4-2x^2c'+\frac {d'}{x}+\frac {2}{x^2}+x$ we obtain, $b'''x+2x-4x^4-2x^2c'''+\frac {d'''}{x}+\frac {2}{x^2}+e''x$ is constant for $x,y,z$ where $b''',c''',d'',e''$ are same for $x,yz$. (well this is easy to show all of $b''\neq b''', c''\neq c''',d''\neq d''', e'\neq e''$. So now we get $-4x^4-2x^2c''''+\frac {d''''}{x}+\frac {2}{x^2}$ , so similarly going on we get a constant polynomial in $x$ for three values with coefficients of $x^k$ are vanishing, for a fixed $k$ (since in each step coefficients of $x^k$ are different for two same degree polynomial in $x$). So keep going on we get $mx$ is constant for $x,y,z$ where $m$ is symmetric in $x,y,z$ and not nonzero either. So $x=y=z$ and so done.

After expanding, we wish to show that \[\left(\sum\limits_{sym} x^3y^4\right) + \left(2\sum\limits_{cyc} x^3y^2z^2\right) \ge \left(\sum\limits_{sym} x^4y^2z\right) + \left(2\sum\limits_{cyc} xy^3z^3\right).\] By putting the difference $\text{LHS} - \text{RHS}$ into Chinese Dumbassing Notation, or by being especially observant, we see that this is equivalent to the "sum" (even though the signs point in opposite directions) of two sum of squares identities: \[\sum\limits_{cyc} x^4z^3 + x^4y^3 \ge \sum\limits_{cyc} x^4y^2z + x^4yz^2 \quad \text{(1)}\]\[\sum\limits_{cyc} 2x^3y^2z^2 \le \sum\limits_{cyc} x^3y^3z + x^3yz^3. \quad \text{(2)}\] By doing a little bit of algebra, we see that $(1)$ is equivalent to \[\sum\limits_{cyc} x^4\left(y^3 + z^3 - y^2z - yz^2\right) \ge 0\]\[\iff \sum\limits_{cyc} x^4\left(y + z\right)\left(y - z\right)^2 \ge 0. \quad \text{(3)}\] Meanwhile, $(2)$ is equivalent to \[0 \le \sum\limits_{cyc} x^3yz\left(y^2 + z^2 - 2yz\right)\]\[\iff 0 \le \sum\limits_{cyc} x^3yz\left(y - z\right)^2. \quad \text{(4)}\] Now, even though their signs point in different directions, we attempt to "add" $(3)$ and $(4).$ We then need to show that \[\sum\limits_{cyc} x^3\left(zx + xy - yz\right)\left(y - z\right)^2 \ge 0.\] Now, by the symmetry of the inequality, we assume WLOG that $z \ge y \ge x.$ Then if we let $S_x = x^3\left(zx + xy - yz\right)$ and denote $S_y$ and $S_z$ similarly, the inequality is equivalent to \[S_x\left(y - z\right)^2 + S_y\left(z - x\right)^2 + S_z\left(x - y\right)^2 \ge 0. \quad \text{(5)}\] We now use a standard sum of squares trick. (See http://www.mit.edu/~evanchen/handouts/SOS_Dumbass/SOS_Dumbass.pdf for more detail) By our ordering assumption, it follows that $y$ is the median of $\{x, y, z\}.$ Therefore \[\left(y - z\right)\left(y - x\right) \le 0\]\[\implies \left(z - x\right)^2 \ge \left(x - y\right)^2 + \left(y - z\right)^2.\] In addition, clearly $S_y = xy + yz - zx \ge yz - zx \ge 0.$ Therefore, we have that \[S_x\left(y - z\right)^2 + S_y\left(z - x\right)^2 + S_z\left(x - y\right)^2 \ge\]\[\ge S_x\left(y - z\right)^2 + S_y\left(\left(x - y\right)^2 + \left(y - z\right)^2\right) + S_z\left(x - y\right)^2\]\[= \left(S_y + S_z\right)\left(x - y\right)^2 + \left(S_x + S_y\right)\left(y - z\right)^2\]\[= yz\left(x - y\right)^2 + xy\left(y - z\right)^2 \ge 0.\] Therefore, it follows that $(5)$ is true, which, upon expansion, is equivalent to the desired result.

Expand the desired inequality as \[\sum_{\mbox{sym}} x^4y^3 + 2xyz[x^2yz+xyz^2+xy^2z]\ge \sum_{\mbox{sym}} x^4y^2z + 2xyz[x^2y^2+y^2z^2+z^2x^2].\]This rearranges to \[\sum_{\mbox{cyc}}x^4[y^3+z^3-y^2z-yz^2] = \sum_{\mbox{cyc}}x^4(y-z)^2(y+z) \ge xyz[(xy-yz)^2+(xy-xz)^2+(xz-yz)^2].\]Rearrange again to \[0\le \sum_{\mbox{cyc}}\left[x^4(y-z)^2(y+z)-x^3yz(y-z)^2\right] = \sum_{\mbox{cyc}}x^3(y-z)^2[xy+xz-yz].\]Let $yz=a, xz=b, xy=c$ to see it is equivalent to prove \[0\le \sum_{\mbox{cyc}}bc\cdot (b-c)^2[b+c-a].\]We are done so long as $a,b,c$ satisfy the triangle inequality. Suppose they do not, so WLOG $a=b+c+t$ for some $t>0$. Then the desired inequality is \[tbc(b-c)^2 \le (b+c+t)c(b+t)^2(2c+t) + (b+c+t)b(c+t)^2(2b+t).\]This rewrites as \[tbc(b-c)^2 \le (b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3].\]In fact, this is trivial because \[(b+c+t)[4b^2c^2+5bct(b+c)+2(b+c)^2t^2+(b+c)t^3]\ge 5(b+c)^2bct > (b+c)^2bct> (b-c)^2bct.\]

SOS Lemma: If $S=S_a(b-c)^2+S_b(a-c)^2+S_c(a-b)^2$ and $S_b,S_a+S_b,S_c+S_b\geq 0$ for $a\geq b\geq c$, then $S\geq 0$. Proof: $(b-a)(b-c)\leq 0\iff (a-c)^2\geq (a-b)^2+(a-c)^2$ hence \[S=S_a(b-c)^2+S_b(a-c)^2+S_c(a-b)^2\geq S_a(b-c)^2+S_b(a-b)^2+S_b(b-c)^2+S_c(a-b)^2\geq 0\] Expanding gives \[\sum_{sym}{x^3y^4}+2x^2y^2z^2\sum{x}\overset{?}{\geq} \sum_{sym}{x^3y}+2\sum{x^2y^2}\]Let $x=\frac{bc}{a}, \ y=\frac{ac}{b}, \ z=\frac{ab}{c}$. If we rewrite the inequality, we get \[\sum_{sym}{ab^4}+2abc\sum{bc}\overset{?}{\geq} \sum_{sym}{a^2b^3}+2abc\sum{a^2}\]\[\iff \sum{ab(a^3+b^3-a^2b-ab^2)}\overset{?}{\geq} 2abc(\sum{a^2}-\sum{bc})=abc[(b-c)^2+(a-c)^2+(a-b)^2]\]\[\iff (b^2c+bc^2-abc)(b-c)^2+(a^2c+ac^2-abc)(a-c)^2+(a^2b+ab^2-abc)(a-b)^2\overset{?}{\geq} 0\]We have $S_a=b^2c+bc^2-abc, \ S_b=a^2c+ac^2-abc, \ S_c=a^2b+ab^2-abc$. WLOG $a\geq b\geq c$. \[S_b=a^2c+ac^2-abc=ac(a-b)+ac^2\geq 0\]\[S_a+S_b=b^2c+bc^2+a^2c+ac^2-2abc=c(a-b)^2+c^2(a+b)\geq 0\]\[S_b+S_c=a^2c+ac^2+a^2b+ab^2-2abc=a(b-c)^2+a^2(b+c)\geq 0\]Which completes the proof by SOS Lemma as desired.$\blacksquare$