Assume that $\delta >0$, and $a,b,c>0$ with $ab=c$.
I am considering the following quadratic form:
$$ q(x_1,x_2,x_3,x_4)=((1+a+b)x_1+bx_2+ax_3-x_4)^2+\delta\left(\frac{x_2^2}{a}+\frac{x_3^2}{b}+\frac{x_4^2}{c}\right), $$
I try to find out whether this quadratic form is positive definit, i.e. whether $q(\vec{x})>0$ for all $\vec{x}=(x_1,\ldots,x_4)\neq (0,0,0,0)$.
I wrote down the corresponding symmetric matrix, i.e., $q(\vec{x})=\vec{x}A\vec{x}^\top$ with $$ A=\begin{pmatrix}(1+a+b)^2 & (1+a+b)\cdot b & (1+a+b)\cdot a & -(1+a+b)\\(1+a+b)\cdot b & b^2+\frac{\delta}{a} & c & -b\\(1+a+b)\cdot a & c & a^2+\frac{\delta}{b} & -a\\-(1+a+b) & -b & -a & 1+\frac{\delta}{c}\end{pmatrix} $$
Obviously, being the sum of two squares, $q\ge 0$ always. The case $q=0$ can only occur when both terms are zero. But then we must have $$ x_2=x_3=x_4=0 $$ from the second term and therefore $$ ((1+a+b)x_1)^2=0 $$ from the first term which implies $x_1=0\,.$