Prove that if three real numbers $a, b, c$ satisfy $a^2+b^2+c^2=ab+bc+ac$, then $a=b=c$.

Question

I don't have much to start the problem, as I have trouble approaching it. What method of solving should I use here?

Answer 1

$$(a^2+b^2+c^2)-(ab+ac+bc) = \frac{1}{2}\left[(a-b)^2+(a-c)^2+(b-c)^2\right]\geq 0 $$ and the middle term equals zero iff $a=b=c$.

Alternative approach: by the Cauchy-Schwarz inequality $$\left|ab+ac+bc\right|\leq \sqrt{a^2+b^2+c^2}\sqrt{b^2+c^2+a^2} = a^2+b^2+c^2$$ and the equality holds iff the vectors $(a,b,c)$ and $(b,c,a)$ are linearly dependent, i.e. iff $a=b=c$.

Answer 2

Hint $$(a-b)^2=a^2-2ab+b^2 \geq 0 $$ and similarly for $(b,c)$ and $(c,a)$.

Answer 3

Finding a change of basis so that the Hessian matrix $H$ is congruent to a diagonal matrix $D$ leads to $$ Q^T D Q = H $$ Sylvester's Law of Inertia says that the diagonal matrix $D$ must have two positive entries and a zero on the diagonal. These need not be the diagonals (except for the zero). $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - \frac{ 1 }{ 2 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) $$ or $$ \color{blue}{ x^2 + y^2 + z^2 - yz - zx - xy = \frac{1}{4} (2x-y-z)^2 + \frac{3}{4} (y-z)^2 } $$

This expression also tells us that the quadratic form is zero only when $x=y=z$

A version without fractions uses $u = x-y,$ $v = y-z,$ then $$ u^2 + uv + v^2 = x^2 + y^2 + z^2 - yz - zx - xy. $$ The conclusion is the same, as $u^2 + uv + v^2$ is strictly positive unless $u=v=0.$

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I put a bunch of information about the algorithm below at reference for linear algebra books that teach reverse Hermite method for symmetric matrices

I call the symmetric matrix of interest $H$ for Hessian:

$$ H = \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) $$

$$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrr} 2 & 0 & - 1 \\ 0 & \frac{ 3 }{ 2 } & - \frac{ 3 }{ 2 } \\ - 1 & - \frac{ 3 }{ 2 } & 2 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrr} 1 & 0 & \frac{ 1 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & - \frac{ 3 }{ 2 } \\ 0 & - \frac{ 3 }{ 2 } & \frac{ 3 }{ 2 } \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 \\ 1 & 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 1 }{ 2 } & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 \\ - \frac{ 1 }{ 2 } & - 1 & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 2 & 0 & 0 \\ 0 & \frac{ 3 }{ 2 } & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } \\ 0 & 1 & - 1 \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 2 & - 1 & - 1 \\ - 1 & 2 & - 1 \\ - 1 & - 1 & 2 \\ \end{array} \right) $$

Answer 4

• Suppose suppose loss of generality that $a\le b\le c$ then $\begin{cases}b=a+u&u\ge 0\\c=b+v&v\ge 0\end{cases}$

$a^2+b^2+c^2-(ab+bc+ca)=u^2+\underbrace{uv}_{\ge 0}+v^2=0\implies u=v=0\implies a=b=c$

• Another interesting substitution is $c=ta+(1-t)b$

$a^2+b^2+c^2-(ab+bc+ca)=(a-b)^2\underbrace{(t^2-t+1)}_{\text{no real root}}=0\implies a=b\implies a=b=c$