Prove inequality involving square roots of trigonometric functions For $N$=even and $x\geq0$, prove that
\begin{eqnarray}
&&\sum^{\frac{N}{2}}_{j=1}\sqrt{x^2-2x\cos\frac{(2j-1)\pi}{N}+1}\geq1+\sum^{\frac{N}{2}-1}_{j=1}\sqrt{x^2-2x\cos\frac{2j\pi}{N}+1}
\end{eqnarray}
where the equality holds for $x=0$.

Update 1: As shown by Y.Guo (see the answer below), the original inequality is equivalent to
\begin{eqnarray}
f(x)\equiv\sum^{\frac{N}{2}}_{j=1}\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{x^2-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^2-2x\cos\frac{2j\pi}{N}+1}}\leq 1.
\end{eqnarray}
It is easy to show that $f(0)=1$ and $f(1)<1$. When $x\geq \cos\frac{\pi}{N}$, each term in the sum is a monotonically decreasing function, which proves the inequality for $x\in[\cos\frac{\pi}{N},\infty)$.
However, numerical tests indicate that $f(x)$ also decreases monotonically on $x\in[0,\cos\frac{\pi}{N}]$. I have no idea how to prove this since some terms in the sum are not monotonical functions on this interval.

Update 2: The inequality for $0<x<1$ can be proved by combining the Perron-Frobenius theorem in matrix theory with the fermion representation of the one-dimensional quantum Ising model.
This can be viewed as a kind of “physical mathematics” in some sense. See the following paper for details:
https://arxiv.org/abs/2001.00511
 A: A Proof  when $x\ge1$
$$
\begin{aligned}
\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}&\ge1+\sum_{j=1}^{\frac{N}{2}-1}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+x&\ge{x}+1+\sum_{j=1}^{\frac{N}{2}-1}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+x&\ge\sum_{j=1}^{\frac{N}{2}}\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}\\
\iff\sum_{j=1}^{\frac{N}{2}}(\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}&-\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1})+x\ge0\\
\end{aligned}
$$
$$
\begin{aligned}
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2x\cos\frac{2j\pi}{N}-2x\cos\frac{(2j-1)\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}+x\ge0
\end{aligned}
$$
$$
\begin{aligned}
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2\cos\frac{2j\pi}{N}-2\cos\frac{(2j-1)\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}+1\ge0\\
\iff\sum_{j=1}^{\frac{N}{2}}&\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{x^{2}-2x\cos\frac{(2j-1)\pi}{N}+1}+\sqrt{x^{2}-2x\cos\frac{2j\pi}{N}+1}}\le1
\end{aligned}
$$
Notice that the left  hand side decreases as $x$ increases when $x\ge1$
($\forall 1\le{j}\le\frac{N}{2}:2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}>0$ and $x^{2}-2x\cos\frac{k\pi}{N}+1$ increases for all $1\le{k}\le{N}$ $x\ge1$)
So if $x=1$ the inequality holds, so does $x\ge1$
When $x=1$
$$
\begin{aligned}
LHS&=\sum_{j=1}^{\frac{N}{2}}\frac{2\cos\frac{(2j-1)\pi}{N}-2\cos\frac{2j\pi}{N}}{\sqrt{2-2\cos\frac{(2j-1)\pi}{N}}+\sqrt{2-2\cos\frac{2j\pi}{N}}}\\
&=\sum_{j=1}^{\frac{N}{2}}\sqrt{2-2\cos\frac{(2j-1)\pi}{N}}-\sqrt{2-2\cos\frac{2j\pi}{N}}\\
&=\sum_{j=1}^{\frac{N}{2}}2(\sin\frac{2j\pi}{2N}-\sin\frac{(2j-1)\pi}{2N})\\
&=\frac{\cos(\frac{\pi}{2N})-\cos(\frac{\pi}{2}+\frac{\pi}{2N})}{\sin(\frac{\pi}{2N})}-\frac{1}{\sin(\frac{\pi}{2N})}\\
&=\frac{\cos(\frac{\pi}{2N})+\sin(\frac{\pi}{2N})-1}{\sin(\frac{\pi}{2N})}\\
&\le1
\end{aligned}
$$
