if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$ Let $x \in \mathbb{R}^*$ and $y \geq \frac{1}{4}$.
Show that if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $  then $y\leq 1$ or $y \geq 4$
I tried the following idea:
$$
2\sqrt{y-\frac{1}{4}}\leq \sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y
$$
which leads to:
$$
y^2 - 4 y +1 \geq 0
$$
then $y \leq 2-\sqrt{3} $ or $ y \geq 2+\sqrt{3} $
 A: Claim: $y \ge 1$.
Proof: For $y \ge \dfrac{1}{4}$ we have: $y = \sqrt{x^2+x+y} + \sqrt{x^2-x+y}\ge \sqrt{x^2+x+\dfrac{1}{4}}+\sqrt{x^2-x+\dfrac{1}{4}}=\left|x+\dfrac{1}{2}\right|+\left|x-\dfrac{1}{2}\right|$. So if $x \le -\dfrac{1}{2}$, then $y \ge -x-\dfrac{1}{2}-x+\dfrac{1}{2}=-2x \ge 1$. If $-\dfrac{1}{2} \le x \le \dfrac{1}{2}$, then $y \ge x+\dfrac{1}{2} - x+\dfrac{1}{2} = 1$. And if $x \ge \dfrac{1}{2}$, then $y \ge x+\dfrac{1}{2}+x-\dfrac{1}{2} = 2x \ge 1$. Thus we have shown that for any real number $x$, $y \ge 1$. Observe this is the range of $y$ as well, and you can choose an $x$ so that $y \ge 4$. Let $x = \dfrac{7}{2}$, then $y \ge \left|\dfrac{7}{2}+\dfrac{1}{2}\right|+\left|\dfrac{7}{2}-\dfrac{1}{2}\right|=4+3=7 > 4$.
Note: The statement that $y \le 1$ or $y \ge 4$ is changed to $y \ge 1$.
A: Let $y=z^2+\frac14$ and
$$y=\sqrt{\left(x+\tfrac12\right)^2+z^2} + \sqrt{\left(\tfrac12-x\right)^2+z^2}$$
Now by triangle/Minkowski inequality,
$$\implies y \geqslant \sqrt{(1)^2+(2z)^2} = \sqrt{4z^2+1}=\sqrt{4y}$$
$\implies y(y-4)\geqslant 0 \implies y\geqslant 4$
Hence among the options provided, $y\leqslant1$ clearly fails. As $(x,y)=(0,4)$ satisfies the original equation, this is indeed the minimum possible.
