If $u^2 \ge -\dfrac{8}{3}$, then $u \ge -\sqrt{\dfrac{8}{3}}$.

Question

If $u^2 \ge -\dfrac{8}{3}$, then $u \ge -\sqrt{\dfrac{8}{3}}$.

Is this the correct convention? I was confused because initially I thought the negative sign would go inside the square root, but then that would lead to imaginary numbers. Thanks.

Answer 1

No, this is wrong.

In general for any real $u$ $$u^2 \geq 0,$$ so your starting inequality is trivially true.

What is in general true is that for any non-negative $a\geq 0$ $$u^2\geq a \iff u \geq \sqrt{a}\ \lor\, -\sqrt{a}\geq u\iff u \in (-\infty,-\sqrt{a}]\ \cup \ [\sqrt{a},+\infty) $$

and conversely

$$u^2\leq a \iff -\sqrt{a}\leq u \leq \sqrt{a} \iff u \in [-\sqrt{a},\sqrt{a}].$$

Answer 2

Can you explain to me how you know $~\displaystyle u^2 \geq a \iff -\sqrt{a} \geq -u ~?$

To attack this entire subject analytically, you need a previous result, from the Field Theory of real numbers.

Suppose that $~a < b.~$ Then:

• For any $~c > 0, ~(c \times a) < (c \times b).~$

• For any $~c < 0, ~(c \times a) > (c \times b).~$

The above result is actually based on some prior ideas from the Field Theory of real numbers:

• $~(a < b) \iff (b - a) > 0.$

• If $~r > 0,~$ and $~s > 0,~$ then $~(r \times s) > 0.$

• If $~r < 0,~$ and $~s < 0,~$ then $~(r \times s) > 0.$

• If exactly one of $~r,s~$ is positive, and the other one is negative, then $~(r \times s) < 0.$

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Now, suppose that $~a > 0,~$ and you are trying to solve the inequality

$$x^2 \geq a. \tag1 $$

The simplest approach will be case work. In order to identify all values of $~x~$ that satisfy the inequality in (1) above, you can split your analysis into the following two cases:

• Case 1 : $~x \geq 0.~$
• Case 2 : $~x < 0.$

Note that by convention, for any $~a > 0,~$ you will always have that $~\sqrt{a} > 0.$

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$\underline{\text{Case 1 :} ~x \geq 0}$

For Case 1, you are to assume that $~x~$ must be $~\geq 0,~$ and then identify all non-negative values of $~x~$ that satisfy the inequality in (1) above.

Suppose that $~\displaystyle 0 \leq x < \sqrt{a}.~$

Then, $~\displaystyle 0 \leq ( ~x \times x ~) < \left( ~x \times \sqrt{a} ~\right) < \left( ~\sqrt{a} \times \sqrt{a} ~\right).~$

Therefore, under the assumption that $~0 \leq x,~$ there can be no satisfying solution for (1) above, where $~\displaystyle x < \sqrt{a}.~$

Alternatively, suppose that $~\displaystyle 0 < \sqrt{a} \leq x.$

Then, $~\displaystyle 0 \leq \left( ~\sqrt{a} \times \sqrt{a} ~\right) \leq \left( ~x \times \sqrt{a} ~\right) \leq ( ~x \times x ~).~$

Therefore, under the Case 1 assumption that $~x \geq 0,~$ the solution to (1) above is represented by $~\displaystyle \sqrt{a} \leq x.$

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$\underline{\text{Case 2 :} ~x < 0}$

For Case 2, you are to assume that $~x~$ must be $~< 0,~$ and then identify all negative values of $~x~$ that satisfy the inequality in (1) above.

Suppose that $~\displaystyle - \sqrt{a} < x < 0.~$

Then, $~\displaystyle \left( ~ -\sqrt{a} \times -\sqrt{a} ~\right) > \left( ~-\sqrt{a} \times x ~\right) > ( ~x \times x ~).~$

Therefore, under the assumption that $~x < 0,~$ there can be no satisfying solution for (1) above, where $~\displaystyle - \sqrt{a} < x.~$

Alternatively, suppose that $~\displaystyle x \leq -\sqrt{a} < 0.$

Then, $~\displaystyle ( ~x \times x) \geq \left( ~x \times \sqrt{a} ~\right) \geq \left( ~\sqrt{a} \times \sqrt{a} ~\right).~$

Therefore, under the Case 2 assumption that $~x < 0,~$ the solution to (1) above is represented by $~\displaystyle x \leq -\sqrt{a}.$

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$\underline{\text{Final Summary}}$

From the Case 1 and Case 2 analysis:

• If $~x \geq 0,~$
  then $~x^2 \geq a \iff x \geq \sqrt{a}.$

• If $~x < 0,~$
  then $~x^2 \geq a \iff x \leq -\sqrt{a}.$

Therefore, the complete set of solutions to (1) above is

$$\{ ~x ~: ~\left( ~x \leq -\sqrt{a} ~\right) ~~\text{or} ~~ \left( ~x \geq \sqrt{a} ~\right) ~\}.$$

Answer 3

NOTE that $\forall a \in \mathbb{R} , a^2 \geq 0$ $ so your first inequality is always true .

$u^2 \geq 0 > -\frac{8}{3 } \implies u \in \mathbb{R} $

If you're still not convinced we can move everything to one side and solve the quadratic inequality $u^2+ \frac{8}{3} \geq0$ which has a negative determinant $D=b^2-4 \cdot a \cdot c = 0^2-4 \cdot 1 \cdot \frac{8}{3}=- \frac{32}{3} < 0$ so the trinomial is always possitive