For $a, b ∈ R$ prove the following inequality: $12a^2 + 36ab + 36b^2 + 7 \geq 18a + 24b$
I suppose it has something to do with squared numbers being positive, so I've tried re-writing the inequality in a couple of ways so that I get squared parenthesis. But I always get stuck at something like: $2[(3a+3b)^2] + 7 \geq 6(a^2 + b^2 +3a + 4b)$.
You are correct that it is completing the square
The matrix identity $Q^T DQ = H$ below reads: $$ 12 \left( x + \frac{3y}{2} - \frac{3}{4} \right)^2 + 9 \left( y +\frac{1}{6} \right)^2 = 12 x^2 +36xy+36y^2 + 7 - 18x - 24 y $$
Algorithm discussed at reference for linear algebra books that teach reverse Hermite method for symmetric matrices
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ P^T H P = D $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ - \frac{ 3 }{ 2 } & 1 & 0 \\ 1 & - \frac{ 1 }{ 6 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 12 & 18 & - 9 \\ 18 & 36 & - 12 \\ - 9 & - 12 & 7 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & - \frac{ 3 }{ 2 } & 1 \\ 0 & 1 & - \frac{ 1 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 12 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ $$ $$
$$ Q^T D Q = H $$
$$\left( \begin{array}{rrr} 1 & 0 & 0 \\ \frac{ 3 }{ 2 } & 1 & 0 \\ - \frac{ 3 }{ 4 } & \frac{ 1 }{ 6 } & 1 \\ \end{array} \right) \left( \begin{array}{rrr} 12 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) \left( \begin{array}{rrr} 1 & \frac{ 3 }{ 2 } & - \frac{ 3 }{ 4 } \\ 0 & 1 & \frac{ 1 }{ 6 } \\ 0 & 0 & 1 \\ \end{array} \right) = \left( \begin{array}{rrr} 12 & 18 & - 9 \\ 18 & 36 & - 12 \\ - 9 & - 12 & 7 \\ \end{array} \right) $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$