Proving Inequalities: Tips and Examples for Solving with Different Methods

[courtrigrad]

Hello all

How would you prove the following:

(a) x + \frac{1}{x} \geq 2, x > 0
(b) x + \frac{1}{x} \leq -2, x < 0
(c) |x+\frac{1}{x}| \geq 2, x\neq 0.

For all of these inequalities would I simply solve for x, or would I have to use things like the triangle inequality of Schwarz's Inequality?

Ok I was also given some harder problems with the same concepts:Prove the following inequalities:
(i) x^2 + xy + y^2\geq 0
(ii) x^{2n} +x^{2n-1}y + x^{2n-2}y^2+ ... + y^{2n} \geq 0
(iii) x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0.

Ok for (i) we can factor \frac{x^3-y^3}{x-y}. If x > y, x < y this expression is positive. For x = y we have 3x^2 which is positive.

For (ii) we have the same thing except \frac{x^{2n+1} - y^{2n+1}}{x-y}.

For (iii) x^4 - 3x^3 + 4x^2 - 3x + 1 \geq 0<br /> \x^4 - 3x^3 + 4x^2 - 3x + 1 = (x-1)^{2}(x^2-x+1)
x^2 - x + 1 = (x-\frac{1}{2})^2 + \frac{3}{4} \geq \frac{3}{4}

Would the above problems be similar to these?

Thanks

 

[arildno]

For your first:
Differentiate x+1/x.
You should be able to show that for x>0, x=1 is a minimum.
Do somewhat analogous for x<0

 

[courtrigrad]

well actually you are not supposed to use differentiation as this is the first chapter in the calculus book. so would i just solve for x?

thanks

 

[arildno]

In that case, assume for x>0, x+\frac{1}{x}&lt;2
Show that this leads to a contradiction.

Make a similar argument in the case x<0

 

[HallsofIvy]

I confess I don't see how differentiating would help.

To solve x+ \frac{1}{x}\ge 2, as long as you know x> 0, you can multiply the entire inequality by x to get x^2+ 1\ge 2x or x^2- 2x+ 1\ge 0 That factors as (x-1)^2\ge 0 which, since a square is always positive, is true for all x.

 

[courtrigrad]

ok thanks a lot

 

[arildno]

I confess I don't see how differentiating would help.

Eeh, define the function: f(x)=x+\frac{1}{x}
Hence, f&#039;(x)=1-\frac{1}{x^{2}}
which shows that x=\pm1 are critical points.
The 2. derivative test shows that x=1 is a local minimum (with f(1)=2), whereas x=-1 is a local maximum f(-1)=-2.

Furthermore, by inspecting the signs of f' on either side of, say, x=1, we may readily conclude that x=1 is a "global" minimum for x>0 (that is, f(x)>=2, x>0).