= Solution
Since every $u_{jj}>0$, the <diagonal matrix> $D$ is invertible. Set
$$
\widehat U=D^{-1}U,\qquad
\widehat u_{ij}=\frac{u_{ij}}{u_{ii}}.
$$
The division is by the diagonal entry of row $i$. Thus $\widehat U$ is an <upper triangular matrix> with diagonal entries one and $A=LD\widehat U$.
To use symmetry without assuming the conclusion, multiply by the <matrix inverse> of $L$ and its <transpose>:
$$
M=L^{-1}AL^{-T}=D\widehat U L^{-T}.
$$
The <matrix> $M$ is symmetric because $A$ is a <symmetric matrix>. It is also an <upper triangular matrix>, since every factor on the right is an <upper triangular matrix>. A symmetric <upper triangular matrix> is diagonal. Its diagonal is exactly $D$, because $\widehat U$ and $L^{-T}$ have diagonal entries one. Hence $M=D$, giving
$$
\boxed{A=LDL^T,\qquad \widehat U=L^T.}
$$
This is the <LDL decomposition>.
Take the positive <square roots> of the diagonal entries:
$$
D^{1/2}=\operatorname{diag}(\sqrt{u_{11}},\ldots,\sqrt{u_{nn}}),\qquad
\widehat L=LD^{1/2}.
$$
The <matrix> $\widehat L$ is a <lower triangular matrix> with positive diagonal, and
$$
\boxed{A=\widehat L\widehat L^T.}
$$
This is a <Cholesky decomposition>. For a nonzero <vector> $x$, the <vector> $\widehat L^Tx$ is nonzero because $\widehat L^T$ is invertible. Therefore
$$
\boxed{x^TAx=(\widehat L^Tx)^T(\widehat L^Tx)=\|\widehat L^Tx\|_2^2>0.}
$$
Here $\|\cdot\|_2$ is the <Euclidean norm>. Thus $A$ is a <positive-definite matrix>.
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