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The question I'm having trouble with is follows:

Suppose you have already proved the proposition that, “If $a$ and $b$ are nonnegative real numbers, then $\frac{a + b}{2} ≥ \sqrt{ab}$.”

a. Explain how you could use this proposition to prove that if a and b are real numbers satisfying the property that $b ≥ 2|a|$, then $b ≥ \sqrt{b^2 − 4a^2}$. Be careful how you match up notation.

b. Use the foregoing proposition and part (a) to prove that if $a$ and $b$ are real numbers with $a < 0$ and $b ≥ 2|a|$, then one of the roots of the equation $ax^2 + bx + a = 0$ is $≤ -\frac{b}{a}.$

I'm stuck on the solution to $b$. (sadly), because the explanation involves dividing an inequality by another inequality and I have no idea how it works.

Here is the solution: solution screenshot

Between $A_1$ and $A_2$ in solution to b., it says "Subtracting $2b$ from both sides of $A_1$ and then dividing by $2a <0$ (from the hypothesis)". I don't know how dividing by $2a <0$ works. I mean, I know how to divide an equation by another equation in systems of equations, but oh god what is this. Please help me

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  • $\begingroup$ Please try to use link LaTeX next time. $\endgroup$ – Jaemin Kim Oct 25 '19 at 4:07
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You shouldn't read that line as "divide by the inequality $2a<0$." Rather, read it as "divide both sides by $2a$, which is negative (so dividing by $2a$ flips the direction of the inequality)."

In some situations, it may be possible to "divide an inequality by another inequality." If $a > b$ and $0 < c < d$, then $\frac{a}{c}>\frac{b}{d}$, and you could perhaps think of this conclusion as being reached by dividing the inequality $a>b$ by the inequality $c < d$. But even in this case it would be clearer to split the argument up into two steps: first note that $\frac{a}{c}>\frac{b}{c}$, and then that $\frac{b}{c}>\frac{b}{d}$.

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Well, think of it as simply dividing by a negative quantity. That's the idea you want to get from seeing $2a<0.$ And actually, it's only in such cases (when one side of a strict inequality is $0$) that you can divide by such inequalities.

In sum, what we've done is to divide by $2a,$ and this reverses the order accordingly.

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