Proofs Involving Inequalities

[turophile]

I'm reviewing some problems from my old calculus textbook, and a couple of them have me stumped. Here's the first one:

Prove that if x > 0 and x^2 > 2, then x^2 > (x/2 + 1/x)^2 > 2. The textbook gives this hint: x/2 > 1/x, Also, (x/2 - 1/x)^2 > 0.

Here's the second one:

Prove that if x >= 0, then x^3 >= 3x - 2. The textbook gives this hint: Write x^3 = x * x^2 and apply the proof of x^2 >= 2x - 1. By the way, I did figure out that last proof (I think). Something along the lines of (x - 1)^2 >= 0 ==> x^2 - 2x + 1 >=0 ==> x^2 >= 2x - 1.

Thanks for any help on these.

 

[daon]

First one: Square out what you have in parentheses, what is your middle term?

Note you only have to prove x > x/2+1/x > sqrt(2). Here's another hint: x = x/2+x/2 > ?.

Second one, you use the hint: x^3=x^2*x >= x(2x-1) = 2x^2-x >= 2(2x-1)-x >= ?

 

[turophile]

OK, I think I have the first proof (thanks to your hints, daon):

1. x > 0 and x^2 > 2 ==> (x^2)/2 > 1 ==> x/2 > 1/x

2. x = x/2 + x/2 > x/2 + 1/x ==> x^2 > (x/2 + 1/x)^2

3. x > 0 ==> (x/2 - 1/x)^2 > 0 ==> (x^2)/4 - 1 + 1/(x^2) > 0 ==> (x^2)/4 + 1/(x^2) > 1 ==> (x^2)/4 + 1/(x^2) + 1 > 2 ==> (x/2 + 1/x)^2 > 2

So x^2 > (x/2 + 1/x)^2 > 2. Does that look right?

Now I'll work on the second one.

 

[turophile]

So here's my stab at the second proof:

1. (x - 1)^2 >= 0 ==> x^2 - 2x + 1 >= 0 ==> x^2 >= 2x - 1

2. x >= 0 ==> x^3 = x^2 * x >= x(2x - 1) = 2x^2 - x >= 2(2x - 1) - x = 4x - 2 - x = 3x - 2

So if x >= 0, x^3 >= 3x - 2.

That looks right, I think. Let me know otherwise. Many thanks, daon.