Find k for Positive Definite Quadratic Form I have two quadratic forms and I need to find $k$ (different $k$ for each possibly) that makes them positive definite. Here are the two:


*

*$Q(y)=5y_1^2+y_2^2+ky_3^2+4y_1y_2=2y_1y_3-2y_2y_3$

*$Q(y)=ky_1^2+ky_2^2+ky_3^2+2y_1y_2+2y_1y_3-2y_2y_3$


What I would like to do is ensure that each expression is always positive for our chosen range of $k$, but I am not supposed to use matrices. Only the quadratic form definition.
My attempt: I was trying to factor the equations in an attempt to get some square terms (which are always positive) and then some other terms that could determine k, but I was unable to work out the algebra.
 A: By matrix for the first we have
$$Q(y)=5y_1^2+y_2^2+ky_3^2+4y_1y_2-2y_1y_3-2y_2y_3=y^TAy$$
$$A=\begin{bmatrix}
5 & 2 & -1\\
2 & 1 & -1\\
-1 & -1 & k\end{bmatrix}$$
now observe that
$$\det(5)=5>0 \quad \det\begin{bmatrix}
5 & 2 \\
2 & 1 \\
\end{bmatrix}=1>0$$
$$\det A=5k+2+2-1-5-4k=k-2>0\iff k>2$$
thus for Sylvester's criteria the quadratic form is definite positive for $k>2$.
For the second
$$Q(y)=ky_1^2+ky_2^2+ky_3^2+2y_1y_2+2y_1y_3-2y_2y_3$$
$$A=\begin{bmatrix}
k & 1 & 1\\
1 & k & -1\\
1 & -1 & k\end{bmatrix}$$
now observe that
$$\det(k)=k>0 \quad \det\begin{bmatrix}
k & 1 \\
1 & k \\
\end{bmatrix}=k^2-1>0 \implies k>1$$
$$\det A=k^3-1-1-k-k-k=k^3-3k-2>0\iff k>2$$
thus for Sylvester's criteria the quadratic form is definite positive for $k>2$
A: For the first we need to find a value of $k$ for which
$$5a^2+b^2+kc^2+4ab-2ac-2bc\geq0$$ or
$$kc^2-2(a+b)c+5a^2+b^2+4ab\geq0$$ for which we need
$k>0$ and since it's a quadratic inequality of $c$, we need also $\Delta\leq0$, which is $$(a+b)^2-k(5a^2+b^2+4ab)\leq0$$ or
$$(5k-1)a^2+2(2k-1)ab+(k-1)b^2\geq0,$$ for which we need
$$(2k-1)^2-(5k-1)(k-1)\leq0$$ or
$$k^2-2k\geq0,$$ which gives $$k\geq2.$$
The work with the second form is the same.
I got that all $k\geq2$ they are valid.
By the way, for the second there is an easy solution:
For $k\geq2$ we obtain:
$$k(a^2+b^2+c^2)+2ab+2ac-2bc=$$
$$=(k-2)(a^2+b^2+c^2)+(a+b)^2+(a+c)^2+(b-c)^2\geq0.$$
I used the following.
For $a\neq0$ we obtain: $$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}\right)=$$
$$=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a^2}\right)=a\left(\left(x+\frac{b}{2a}\right)^2-\frac{\Delta}{4a^2}\right).$$ If we want that $ax^2+bx+c\geq0$ for all real $x$ then we need $a>0$ and $\Delta\leq0.$
