MHB Prove Positive Fraction + Inverse is ≥2

AI Thread Summary
The discussion centers on proving that the sum of a positive fraction and its inverse is greater than or equal to two. The proof demonstrates that the expression can be rewritten and simplified to show that (a-b)² is always non-negative, confirming the inequality. Participants question the validity of taking the square root of both sides of the inequality, noting that this could lead to incorrect conclusions if a is less than b. The distinction between squaring and square rooting in inequalities is emphasized, particularly the importance of considering absolute values. Overall, the proof is validated, but caution is advised regarding the manipulation of inequalities.
IHateFactorial
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Problem: Prove that any positive fraction plus its inverse is greater than or equal to two.

Proof:

$$\frac{a}{b}+\frac{b}{a}\ge2$$

$$\frac{a^2+b^2}{ab}\ge2$$

$${a^2+b^2}\ge2ab$$

$$a^2+b^2 - 2ab\ge0$$

$$a^2 - 2ab + b^2\ge0$$

$$(a-b)^2\ge0$$

This is true for all a and b:

Case 1:
$$a>b\therefore a-b>0; (a-b)^2>0$$

Case 2:
$$a=b\therefore a-b=0; (a-b)^2=0$$

Case 3:
$$a<b\therefore a-b<0; (a-b)^2>0$$

Tadaaaa!?

Is this right?

If so, one question:

$$(a-b)^2\ge0$$

Can't we just square root both sides and get:

$$a-b\ge0$$

I imagine the rules for squares and square roots are different when working with inequalities... HOW are they different?
 
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IHateFactorial said:
Problem: Prove that any positive fraction plus its inverse is greater than or equal to two.

Proof:

$$\frac{a}{b}+\frac{b}{a}\ge2$$

$$\frac{a^2+b^2}{ab}\ge2$$

$${a^2+b^2}\ge2ab$$

$$a^2+b^2 - 2ab\ge0$$

$$a^2 - 2ab + b^2\ge0$$

$$(a-b)^2\ge0$$

I would have started from the bottom and worked my way up. As it is, you're starting out by assuming what you want to prove is true.

IHateFactorial said:
... one question:

$$(a-b)^2\ge0$$

Can't we just square root both sides and get:

$$a-b\ge0$$

What if $b>a$? In general, $\sqrt{(x-y)^2}=|x-y|$.
 
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