How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?

Question

How can the following inequation be proven?

$$a^2 + b^2 + c^2 \ge ab + bc + ca$$

Answer 1

Try $(a-b)^2+(b-c)^2+(c-a)^2 \ge0$

Compute lhs, divide by two and rearrange.

Answer 2

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.

Answer 3

$$\sum_{cyc}(a^2-ab)=\frac{1}{2}\sum_{cyc}(a^2-2ab+b^2)=\frac{1}{2}\sum_{cyc}(a-b)^2\geq0$$

Answer 4

$$q := a^2 + b^2 + c^2 - ab - bc - ca = \frac12 \begin{bmatrix} a\\ b \\ c\end{bmatrix}^\top \underbrace{\begin{bmatrix} 2 & -1 & -1\\ -1 & 2 & -1 \\ -1 & -1 & 2\end{bmatrix}}_{=: {\rm L}} \begin{bmatrix} a\\ b \\ c\end{bmatrix}$$

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 - ab - ac - b*c
  2          2                2

o3 = a  - ab + b  - ac - 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

Answer 5

Schur inequality where $r= 0.$

Answer 6

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!

Answer 7

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.

Answer 8

$a^2+b^2+c^2-ab-bc-ca\\ =(a, b, c)\begin{pmatrix}1&-1/2&-1/2\\-1/2& 1&-1/2\\-1/2&-1/2&1\end{pmatrix}\left(\begin{array}{c}a\\b\\c\end{array}\right)$

$=(a, b, c)A\begin{pmatrix}a\\b\\c\end{pmatrix}$

It is sufficient to prove $A$ is a positive semi-definite.

It follows from that the minor determinant of $A$ is one of $1,\dfrac{3}{4},0$

Answer 9

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 $a^2 + b^2 + c^2 \ge ab + bc + ca$

Answer 10

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

Answer 11

Another way to approach this question would be to use the AM-GM Inequality. By AM-GM Inequality,

$ \begin{split} \frac{a^3+b^3+c^3}{3} \geq \sqrt[3]{a^3b^3c^3} & \implies a^3+b^3+c^3 \geq 3abc \\ & \implies a^3+b^3+c^3 - 3abc \geq0 \\ & \implies (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \geq 0 \\ & \implies a^2+b^2+c^2 \geq ab+bc+ca \end{split} $

Answer 12

Obtained $$\sum a^{2}- \sum ab= \left ( c+ a- 2b \right )^{2}+ 3\left ( a- b \right )\left ( b- c \right )$$ by assuming $b:=\mathsf{med}\left ( a, b, c \right ).$

Answer 13

Obtained $$\sum a^{2}- \sum ab= \sum a\left ( a- b \right )\geq 0$$ by assuming $a\geq b\geq 0\geq c.$

Answer 14

Seems like various answers are being posted, so I’m adding one more.

This can be solved with Rearrangement Inequality, as mentioned in the OP’s comment.

---

Theorem 1. Greedy Algorithm

Greedy algorithm is a short-sighted algorithm to gain the best benefit. For example, we have three boxes with \$$20$, \$$50$, \$$100$ bills in each. We can take $3$, $4$, $5$ bills in three different boxes. To make the best benefit, you’ll take $5$ bills from \$$100$ box, $4$ from \$$50$, $3$ from \$$20$.

Of course, this algorithm sometimes(or often) have a failure, as the counterexample given below.

\begin{align} &7 \begin{cases} &20 \begin{cases} &10 \\ \\\ &11 \end{cases}\\ \\ & 19 \begin{cases} &14 \\ \\ &100\end{cases} \end{cases} \end{align}

To make the highest sum, you’ll gotta choose $7-19-100$. But the greedy algorithm chooses $7-20-11$, fails to achieve the goal. This is why the algorithm is called short-sighted.

---

Theorem 2. Rearrangement Inequality

Anyway, based on the algorithm above, we can make some successful inequality called Rearrangement Inequality.

The inequality is written as below:

$$a_1\le a_2\le \cdots\le a_n, b_1\le b_2\le \cdots\le b_n. \\ \sum_{i=1}^{n}a_ib_{n+1-i} \le \sum_{i=1}^{n}a_id_i \le \sum_{i=1}^{n} a_ib_i. \\ (d_i: \text{Rearrangement of sequence } b_i.)$$

I love proving this with induction.

Suppose that there exists a specific rearrangement of $b_i$(not equal to $b_i$) defined as $d_i$ which has the maximum of the sum of $a_id_i$.

Now, there exists a pair of distinct numbers $(k, l)$ which satisfies $k<l, d_k>d_l$.

So define new rearrangement $d’_i$ as:

$$d’_i=\begin{cases} & d_i & (i\neq k, i\neq l) \\ & d_k & (i=l) \\ & d_l & (i=k) \end{cases}$$

Then we have:

$$\sum_{i=1}^{n} a_id_i-\sum_{i=1}^{n} a_id’_i = a_kd_k+a_ld_l-a_kd_l-a_ld_k=(a_k-a_l)(d_k-d_l)<0. \\ \therefore \sum_{i=1}^{n} a_id_i < \sum_{i=1}^{n} a_id’_i.$$

So it’s a contradiction by the definition of sequence $d_i$.

ISW we can prove the minimum value.

---

Conclusion

This post seems to be needlessly long, but I just wanted to elaborate on the comment.

To conclude, without loss of generality let $a\leq b\leq c$.

Now, define $a_1=b_1=a, a_2=b_2=b, a_3=b_3=c$.

Applying the Rearrangement Inequality, we get:

$$a_1b_3+a_2b_2+a_3b_1 \leq a_1b_2+a_2b_3+a_3b_1 \leq a_1b_1+a_2b_2+a_3b_3$$

, which can be rewritten as:

$$ac+bb+ca\leq ab+bc+ca\leq aa+bb+cc$$

. So we have $ab+bc+ca\leq a^2+b^2+c^2$, and we are finally done!