Prove Positive Fraction + Inverse is ≥2

[IHateFactorial]

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?

 

[Greg]

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.

... 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, $\sqrt{(x-y)^2}=|x-y|$.