<s>[INST] 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} $
 [/INST] 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.
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