# Thread: How do you prove this inequality?

1. ## How do you prove this inequality?

Prove that a/b + b/a is greater or equal to 2, give that a and b >0. Your help is appreciated.

Love,
numbertheoryboy

2. ## Re: How do you prove this inequality?

Originally Posted by numbertheoryboy
Prove that a/b + b/a is greater or equal to 2, give that a and b >0. Your help is appreciated.

Love,
numbertheoryboy
consider $f(x)=\dfrac{a}{x}+\dfrac{x}{a}$ and find the extrema

Spoiler:
$\dfrac{df}{dx}(x)=-\dfrac{a}{x^2}+\dfrac{1}{a}$

$-\dfrac{a}{x^2}+\dfrac{1}{a}=0 \Rightarrow x=a$

determine whether it's a minima or maxima

$\dfrac{d^2f}{{dx}^2}=\dfrac{2a}{x^3} \left |_{x=a}\right. =\dfrac{2a}{a^3}=\dfrac{2}{a^2}>0$

so we have a minimum of $f(x)$ at $x=a$

$f(a)=\dfrac{a}{a}+\dfrac{a}{a}=2 \Rightarrow$

$f(x) \geq 2~~\forall x >0$

3. ## Re: How do you prove this inequality?

Originally Posted by numbertheoryboy
Prove that a/b + b/a is greater or equal to 2, give that a and b >0. Your help is appreciated.
Note that $\\a^2+b^2\ge 2ab\\\dfrac{a^2+b^2}{ab}\ge 2\\\dfrac{a^2}{ab}+\dfrac{b^2}{ab}\ge 2\\\dfrac{a}{b}+\dfrac{b}{a}\ge 2$.

4. ## Re: How do you prove this inequality?

Plato has given you a very elegant proof. It depends on the proposition that $a^2 + b^2 \ge 2ab.$

If that is a theorem that has already been proved, that is sufficient for the proof to be valid. However, how do you determine that such a previously demonstrated theorem is relevant to your proof, and if that theorem has not previously been shown, why would you even dream that proving it might be relevant to your task?

A very useful method (I believe attributed to the Greek mathematician Proclus) for finding proofs is to assume that what is to be proved is true and deducing things from there. Let's see it in action.

$\dfrac{a}{b} + \dfrac{b}{a} \ge 2.$ Assuming true what is to be proved.

$\implies \dfrac{a}{b} * \dfrac{a}{a} + \dfrac{b}{a} *\dfrac{b}{b} \ge 2 \implies \dfrac{a^2}{ab} + \dfrac{b^2}{ab} \ge 2 \implies \dfrac{a^2 + b^2}{ab} \ge 2.$ Experimenting with a common denominator.

$\implies a^2 + b^2 \ge 2ab.$ This follows because $a,\ b >0 \implies ab > 0.$

Now if this is a previously proved theorem, your proof involves reversing the steps. Otherwise you keep going.

$a^2 + b^2 \ge 2ab \implies a^2 - 2ab + b^2 \ge 0 \implies (a - b)^2 \ge 0.$

Provided a and b are real numbers, the square of their difference is certainly non-negative. So my proof would show the logic in reverse and look like

$a,\ b \in \mathbb R\ and\ a > 0 < b.$ Given. Notice that this proof is valid only for positive real numbers.

$(a - b) \in \mathbb R.$

$(a - b)^2 \ge 0.$

$a^2 - 2ab + b^2 \ge 0.$

$a^2 + b^2 \ge 2ab.$ From here I would be copying Plato.

My purpose is not to show a different proof from Plato's. My purpose is to show Proclus's method for finding proofs. What is tricky about that method is you have to be sure each step is reversible. It frequently involves experimenting with different conclusions derived from assuming what is to be proved is true. It does not always work, but it has helped me with many, many proofs.