= Solution
{c}
<summation by parts>, with $S(x)=\sum_{n\leq x}\chi(n)$, gives for an <integer> $N$
$$
\sum_{n\leq N}\chi(n)n^{-s}
=S(N)N^{-s}+s\int_1^N S(x)x^{-s-1}\,dx.
$$
The boundary term tends to zero when $\operatorname{Re}s>0$ because $S$ is bounded. In the original region of <absolute convergence> this proves
$$
\boxed{L(s)=s\int_1^\infty S(x)x^{-s-1}\,dx.}
$$
For a <compact subset> of $\operatorname{Re}s>0$, choose $\delta>0$ below all its real parts and bound $|s|$ there. The integral tail is bounded by a constant times $\int_N^\infty x^{-1-\delta}dx$, uniformly on that <compact set>. Its truncated integrals are <holomorphic function> in $s$, so <locally uniform convergence> proves a <analytic continuation> throughout $\operatorname{Re}s>0$. This is <holomorphy of a Dirichlet series from bounded partial sums>.
The <Euler product>, valid for $\operatorname{Re}s>1$, is
$$
L(s)=\prod_p(1-\chi(p)p^{-s})^{-1}
=\prod_{p\equiv1\ (4)}(1-p^{-s})^{-1}
\prod_{p\equiv3\ (4)}(1+p^{-s})^{-1}.
$$
Expand the logarithms of these absolutely convergent <Euler products>. Writing $\theta=mt\log p$, the logarithm of the product in the desired inequality is
$$
\sum_p\sum_{m\geq1}\frac{
3+4\chi(p)^m\cos\theta+\cos2\theta}{m p^{m\sigma}}.
$$
For odd $p$, $\chi(p)^m$ is $1$ or $-1$. In the first case the numerator is the given nonnegative trigonometric <polynomial>. In the second replace $\theta$ by $\theta+\pi$, which changes the sign of $\cos\theta$ and leaves $\cos2\theta$ unchanged. At $p=2$ the numerator is $3+\cos2\theta\geq2$. Every summand is nonnegative. Exponentiation proves
$$
\boxed{\zeta(\sigma)^3|L(\sigma+it)|^4
|\zeta(\sigma+2it)|\geq1.}
$$
This is the <three-four-one inequality for Euler products> applied to the real character modulo four.
Suppose $t\neq0$ and $L(1+it)=0$. Holomorphy gives $L(\sigma+it)=O(\sigma-1)$ as $\sigma\downarrow1$. The only pole of the <Riemann zeta function> is at one, so $\zeta(\sigma+2it)$ remains bounded, while $\zeta(\sigma)^3=O((\sigma-1)^{-3})$. The displayed product is then $O(\sigma-1)$, contradicting its lower bound. Hence \b[$L(1+it)\neq0$ for $t\neq0$].
At $t=0$ the last zeta factor also has a pole, so that contradiction no longer follows from the inequality. Instead evaluate the convergent alternating series:
$$
L(1)=1-\frac13+\frac15-\cdots
=\int_0^1\frac{dx}{1+x^2}
=\boxed{\frac\pi4\neq0}.
$$
For rigor, integrate the finite geometric identity for $(1+x^2)^{-1}$; the remainder has absolute integral at most $1/(2N+3)$ and tends to zero. <summation by parts> above identifies this value with the <analytic continuation> at one.
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