= Solution
Consider the <orientation-preserving affine group of the real line>, realized as
$$
G=\left\{\begin{pmatrix}a&b\\0&1\end{pmatrix}:a>0,\ b\in\mathbb R\right\}.
$$
The coordinates $(\log a,b)$ identify it smoothly with $\mathbb R^2$, so this <Lie group> is connected. Choose the <Lie algebra> element $E$ and the group element $D$:
$$
E=\begin{pmatrix}0&1\\0&0\end{pmatrix},\qquad D=\begin{pmatrix}2&0\\0&1\end{pmatrix}\in G.
$$
Direct <matrix multiplication> gives $\operatorname{Ad}_D E=DED^{-1}=2E$.
If a <bi-invariant Riemannian metric> existed, every conjugation map $x\mapsto D xD^{-1}$ would be an <isometry> fixing the identity. Its <differential> there, the <Adjoint representation of a Lie group>, would preserve the <positive-definite> <inner product> at the identity. Thus
$$
\|E\|^2=\|\operatorname{Ad}_D E\|^2=\|2E\|^2=4\|E\|^2,
$$
a contradiction since $E\ne0$. Therefore \b[this connected Lie group admits no bi-invariant Riemannian metric]. This is the <adjoint dilation obstruction to a bi-invariant Riemannian metric>, an elementary alternative permitted by the question. The positive definite <Riemannian metric> interpretation matters; no claim about indefinite metrics is needed.
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