How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$? How can the following inequation be proven?
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
 A: Try $(a-b)^2+(b-c)^2+(c-a)^2 \ge0$ 
Compute lhs, divide by two and rearrange.
A: $$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$$
A: $$q := a^2 + b^2 + c^2 - ab - bc - ca = \frac12 \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \underbrace{\begin{bmatrix} 2 & -1 & -1\\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}}_{=: {\rm L}} \begin{bmatrix} a\\ b \\ c\end{bmatrix}$$
where matrix $\rm L$ is the Laplacian of the cycle graph with $3$ vertices, whose (signed) incidence matrix is
$${\rm C} = \begin{bmatrix} -1 & \color{blue}{1} & 0\\ 0 & -1 & \color{blue}{1} \\ \color{blue}{1} & 0 & -1\end{bmatrix}$$
Since $\rm L = C^\top C$, we obtain the following sum of squares (SOS) decomposition
$$2q = (\color{blue}{b} - a)^2 + (\color{blue}{c} - b)^2 + (\color{blue}{a} - c)^2 \geq 0$$
which is the SOS decomposition proposed by Mark Bennet. Since matrix $\rm L$ is rank-$2$, a terser SOS decomposition with only $2$ terms can easily be found — say, via the Cholesky decomposition.
Using Macaulay2,
Macaulay2, version 1.16
with packages: ConwayPolynomials, Elimination, IntegralClosure, InverseSystems, LLLBases, MinimalPrimes, PrimaryDecomposition, ReesAlgebra, TangentCone, Truncations

i1 : needsPackage( "SumsOfSquares" );
--loading configuration for package "NumericalAlgebraicGeometry" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-NumericalAlgebraicGeometry.m2
--loading configuration for package "Bertini" from file /Users/rodrigo/Library/Application Support/Macaulay2/init-Bertini.m2
--warning: symbol "Verbosity" in MinimalPrimes.Dictionary is shadowed by a symbol in SemidefiniteProgramming.Dictionary
--  use the synonym MinimalPrimes$Verbosity

i2 : R = QQ[a,b,c];

i3 : q = a^2 + b^2 + c^2 - a*b - a*c - b*c 

      2          2                2
o3 = a  - a*b + b  - a*c - b*c + c

o3 : R

i4 : sosPoly solveSOS q

             1    1  2    3        2
o4 = (1)(a - -b - -c)  + (-)(b - c)
             2    2       4

o4 : SOSPoly

i5 : tex o4

o5 = $\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right
     \},\,\texttt{ring}\,\Rightarrow\,R\right\}$

In $\TeX$,
$$\texttt{SOSPoly}\left\{\texttt{coefficients}\,\Rightarrow\,\left\{1,\,\frac{3}{4}\right\},\,\texttt{generators}\,\Rightarrow\,\left\{a-\frac{1}{2}\,b-\frac{1}{2}\,c,\,b-c\right\},\,\texttt{ring}\,\Rightarrow\,R\right\}$$

polynomials sum-of-squares-method macaulay2 graph-laplacian
A: Schur inequality where $r= 0.$
A: This is a specific form of Cauchy-Schwarz inequality.
Let $x = (a, b, c)$ and $y = (b, c, a)$ as vectors. 
The inequality is $ | \left< x,y \right>| \le \|x\|\|y\|. $ with standard inner product definition.  One neat trick to prove this is using an auxilary parameter $t,$ and expanding $$ \| x+ty \|^2 = 
\left< x+ty,x+ty \right> =  \|x\|^2 + 2 \left< x,y \right>t +\|y\|^2t^2.$$  We know, this being a square, is non-negative.  Therefore, the discriminant of the polynomial in $t$ is less or equal to zero.  Which is $\left< x,y \right>^2 - (\|x\|\|y\|)^2 \le 0.$  Substituting the values for $x$ and $y$ will do the job.
A: If $\ c> a ,a^2+c^2 \gt 2ac $, because $\ (a-c)^2 \gt 0$
If $\ c>b>a ,c^2+b^2/2+a^2/2 \gt ac+bc $ and $\ b^2/2+a^2/2 \gt ab $, sum of them $\ a^2+b^2+c^2 \gt ab+bc+ac $
If $\ c=b \gt a $ or $\ a=b=c $, it can be solved with same logic.
A: $a^2+b^2+c^2-ab-bc-ca\\ 
=(a, b, c)\begin{pmatrix}1&-1/2&-1/2\\-1/2& 1&-1/2\\-1/2&-1/2&1\end{pmatrix}\left(\begin{array}{c}a\\b\\c\end{array}\right)$
$=(a, b, c)A\begin{pmatrix}a\\b\\c\end{pmatrix}$
It is sufficient to prove $A$ is a positive semi-definite.
It follows from that the minor determinant of $A$ is one of $1,\dfrac{3}{4},0$
A: for $a,b,c>=0$
we know $(a-b-c)^2>=0$ 
i.e $a^2 + b^2 + c^2 - 2 (ab + bc + ca) \ge 0$
i.e $a^2 + b^2 + c^2 \ge ab + bc + ca$
A: This inequality can be solved by simple algebra
we have the equation
$$a^2+b^2+c^2 >= ab+ac+cb$$
multiply and divide 2 on both sides
$$\frac{2}{2}(a^2+b^2+c^2) >= \frac{2}{2}(ab+ac+cb)$$
putting the left side equation on the right side
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
then factorize and multiply by 2 on both the sides
$$\frac{2a^2+2b^2+2c^2 - 2(ab+ac+cb)}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 0$$
$$2*\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2} >= 2*0$$
$$(a-b)^2+(b-c)^2+(c-a)^2 >= 0$$
So the above equation is zero when $a=b=c$
So the above inequality is proved
A: From Cauchy-Schwarz 
$ab+bc+ac=\sqrt{a^2}\sqrt{b^2}+\sqrt{b^2}\sqrt{c^2}+\sqrt{a^2}\sqrt{c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
Moving on;
$ab+bc+ac \leq a^2+b^2+c^2$
Done!
A: Obtained
$$\sum a^{2}- \sum ab= \left ( c+ a- 2b \right )^{2}+ 3\left ( a- b \right )\left ( b- c \right )$$
by assuming $b:=\mathsf{med}\left ( a, b, c \right ).$
A: Obtained
$$\sum a^{2}- \sum ab= \sum a\left ( a- b \right )\geq 0$$
by assuming $a\geq b\geq 0\geq c.$
