I have this system of inequalities

$$ \begin{cases} y^2-3 \geq 0\\ 16y^4-96y^2 \geq 0 \end{cases} $$

the solution for the first inequality is $y\leq -\sqrt{3}$ or $y\geq \sqrt{3}$ and the solution for the second inequality is $-\sqrt{6} \leq y \leq \sqrt{6}$. Then for my result the solution for the system is

$-\sqrt{6} \leq y \leq -\sqrt{3}$ or $\sqrt{3} \leq y \leq \sqrt{6}$

According with this I cannot understand why the range of this function $y = \sqrt{x} + \sqrt{3 -x}$ where the system of inequalities is derived it is only $\sqrt{3} \leq y \leq \sqrt{6}$

  • $\begingroup$ Be careful! To satisfy both conditions, you have to take the intersection of what satisfy condition 1 and what satisfy condition 2! $\endgroup$ May 13, 2016 at 18:05
  • $\begingroup$ Hey @LuísaBorsato thanks to replay. Maybe I wrong but the intersection (where the system is satisfied for y) is not -sqrt(6) =< y =< -sqrt(3) or sqrt(3) =< y =< sqrt(6)? thanks in advance $\endgroup$
    – Gianni
    May 13, 2016 at 18:09
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    $\begingroup$ I wrote an answer! =) Hope to help! $\endgroup$ May 13, 2016 at 18:12
  • $\begingroup$ Because of the properties of square roots, $y\ge 0$. By arithmetic-quadratic mean, $y\le 2·\sqrt{\frac{x+(3-x)}2}=\sqrt{6}$ and this inequality is sharp at $x=1.5$. By $y^2=3+2\sqrt{x}\sqrt{3-x}$ you get $y\ge\sqrt{3}$ which again is sharp at $x=0$ and $x=3$. $\endgroup$ May 13, 2016 at 18:16

1 Answer 1


$$ \begin{cases} y^2-3 \geq 0\\ 16y^4-96y^2 \geq 0 \end{cases} $$

To satisfy condition 1

$y^2 - 3 \geq 0 \iff y^2 \geq 3 \iff -\sqrt{3} \geq y$ or $y \geq \sqrt{3}$

To satisfy condition 2:

$16y^4-96y^2 \geq 0 \iff -\sqrt{6} \leq y \leq \sqrt{6}$

Then, to satisfy both conditions

$y \in (-\infty, -\sqrt{3})\cap(\sqrt{3}, +\infty)\cap(-\sqrt{6}, \sqrt{6}) = (-\sqrt{6}, -\sqrt{3}) \cup (\sqrt{3}, \sqrt{6})$

  • $\begingroup$ @thanks! but I agree (sorry for my English form). Now I don't understand why the range of the function is only for [sqrt(3),sqrt(6)] when the system admit also [-sqrt(3),-sqrt(6)] see math.stackexchange.com/questions/1783925/… $\endgroup$
    – Gianni
    May 13, 2016 at 18:14
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    $\begingroup$ Oh! That's because one of the terms of your function is $\sqrt{x}$ and if $x < 0$, $\sqrt{x}$ is not in $\mathbb{R}$ $\endgroup$ May 13, 2016 at 18:16
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    $\begingroup$ arrrrghhhhh You are right!!!!!!!!!! the system in the equation is satisfied for 0 =< x =< 3!!!! $\endgroup$
    – Gianni
    May 13, 2016 at 18:19

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