Prove Positive Fraction + Inverse is ≥2

[IHateFactorial]

Problem: Prove that any positive fraction plus its inverse is greater than or equal to two.

Proof:

\(\displaystyle \frac{a}{b}+\frac{b}{a}\ge2\)

\(\displaystyle \frac{a^2+b^2}{ab}\ge2\)

\(\displaystyle {a^2+b^2}\ge2ab\)

\(\displaystyle a^2+b^2 - 2ab\ge0\)

\(\displaystyle a^2 - 2ab + b^2\ge0\)

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

This is true for all a and b:

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

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

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

Tadaaaa!?

Is this right?

If so, one question:

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

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

\(\displaystyle a-b\ge0\)

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:

\(\displaystyle \frac{a}{b}+\frac{b}{a}\ge2\)

\(\displaystyle \frac{a^2+b^2}{ab}\ge2\)

\(\displaystyle {a^2+b^2}\ge2ab\)

\(\displaystyle a^2+b^2 - 2ab\ge0\)

\(\displaystyle a^2 - 2ab + b^2\ge0\)

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

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:

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

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

\(\displaystyle a-b\ge0\)

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