prove inequality for every non zero real-number

Question

I need your help... I need to prove the following:

For non-zero $a,b,c,d \in \mathbb{R}$,

$4a^2 + 4b^2 + 4c^2 + 4d^2 + 4ab + 4ac + 4bd + 4cd + 2ad + 2bc > 0$

I know it is true since the inequality came from the fact that \begin{bmatrix} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2\\ 1 & 2 & 2 & 4 \end{bmatrix} is positive definite matrix( matrix A is a positive definite Matrix if and only if for every non-zero vector $x$, $$x^TAx > 0$$ )

I tried to use the following absolute inequality

For $a,b,c,d \in \mathbb{R},$

$$ a^2 + b^2 + c^2 + d^2 \geq ab + ac + bd + cd$$

However, it doesn't work well. If you have some beautiful idea to solve it, I'd appreciate some help.

Answer 1

\begin{eqnarray*} (a+b+c+d)^2+(a+b)^2+(a+c)^2+(b+d)^2+(c+d)^2+a^2+b^2+c^2+d^2 >0. \end{eqnarray*}

Answer 2

If you have trouble doing "completing the square repeatedly" as Donald did, there is an algorithm that will find a (rational) matrix $P$ and diagonal $D$ such that your original symmetric matrix $H$ leads to $P^T HP = D.$ Also, $\det P = \pm 1,$ so we also find $Q = P^{-1}$ such that $Q^TDQ = H.$ Finally, by Sylvester's Law of Inertia, the eigenvalues of $H$ have the same $\pm$ signs as the eigenvalues of $D.$

As you are worried about positivity, let me emphasize that $$ x^T H x = X^T Q^T D Q x = (Qx)^T D (Qx) $$ If we introduce a new vector name, letting $y = Qx,$ then we have written $$x^T H x = y^T D y \; .$$ Furthermore, $Q$ is nonsingular, whenever $x$ is nonzero then $y$ is also nonzero.

Final outcomes of the algorithm: $$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 0 & 1 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 0 & 1 & 0 \\ \frac{ 1 }{ 4 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) $$

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Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) $$ $$ D_0 = H $$ $$ E_j^T D_{j-1} E_j = D_j $$ $$ P_{j-1} E_j = P_j $$ $$ E_j^{-1} Q_{j-1} = Q_j $$ $$ P_j Q_j = Q_j P_j = I $$ $$ P_j^T H P_j = D_j $$ $$ Q_j^T D_j Q_j = H $$

$$ H = \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) $$

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$$ E_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{1} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{1} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{1} = \left( \begin{array}{rrrr} 4 & 0 & 2 & 1 \\ 0 & 3 & 0 & \frac{ 3 }{ 2 } \\ 2 & 0 & 4 & 2 \\ 1 & \frac{ 3 }{ 2 } & 2 & 4 \\ \end{array} \right) $$

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$$ E_{2} = \left( \begin{array}{rrrr} 1 & 0 & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{2} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{2} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{2} = \left( \begin{array}{rrrr} 4 & 0 & 0 & 1 \\ 0 & 3 & 0 & \frac{ 3 }{ 2 } \\ 0 & 0 & 3 & \frac{ 3 }{ 2 } \\ 1 & \frac{ 3 }{ 2 } & \frac{ 3 }{ 2 } & 4 \\ \end{array} \right) $$

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$$ E_{3} = \left( \begin{array}{rrrr} 1 & 0 & 0 & - \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{3} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{3} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{3} = \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & \frac{ 3 }{ 2 } \\ 0 & 0 & 3 & \frac{ 3 }{ 2 } \\ 0 & \frac{ 3 }{ 2 } & \frac{ 3 }{ 2 } & \frac{ 15 }{ 4 } \\ \end{array} \right) $$

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$$ E_{4} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{4} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 0 \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{4} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{4} = \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & \frac{ 3 }{ 2 } \\ 0 & 0 & \frac{ 3 }{ 2 } & 3 \\ \end{array} \right) $$

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$$ E_{5} = \left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$ $$ P_{5} = \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; Q_{5} = \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) , \; \; \; D_{5} = \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) $$

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$$ P^T H P = D $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ - \frac{ 1 }{ 2 } & 0 & 1 & 0 \\ \frac{ 1 }{ 4 } & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & - \frac{ 1 }{ 2 } & - \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & - \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) $$ $$ Q^T D Q = H $$ $$\left( \begin{array}{rrrr} 1 & 0 & 0 & 0 \\ \frac{ 1 }{ 2 } & 1 & 0 & 0 \\ \frac{ 1 }{ 2 } & 0 & 1 & 0 \\ \frac{ 1 }{ 4 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & 1 \\ \end{array} \right) \left( \begin{array}{rrrr} 4 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & \frac{ 9 }{ 4 } \\ \end{array} \right) \left( \begin{array}{rrrr} 1 & \frac{ 1 }{ 2 } & \frac{ 1 }{ 2 } & \frac{ 1 }{ 4 } \\ 0 & 1 & 0 & \frac{ 1 }{ 2 } \\ 0 & 0 & 1 & \frac{ 1 }{ 2 } \\ 0 & 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrrr} 4 & 2 & 2 & 1 \\ 2 & 4 & 1 & 2 \\ 2 & 1 & 4 & 2 \\ 1 & 2 & 2 & 4 \\ \end{array} \right) $$