Proving inequalities - Does induction work?

[ibc]

Proving inequalities - Does induction work??

Homework Statement

prove that, for a,b,c>0, a+b+c=1, 1/a+1/b+1/c≥9

Homework Equations

it says that i might want to use the fact that for all X=/=0, X+1/X ≥ 2

The Attempt at a Solution

using the tip I could make it:
a+1/a+b+1/b+c+1/c ≥ 10
but that's as far as I got.

Homework Statement

prove using induction (or anything else):
|sin(nx)|≤n|sin(x)|
for natural n

Homework Equations

The Attempt at a Solution

well it's true for n=1
after using the induction assumption I made it so I have to prove that:
|sin(x(n+1))|≤|sin(nx)| + |sin(x)|
now I'm suck, I don't see how I could use trigonometry equivalences since they all give me cosins and sins*cosins and stuff like that, and these absolute values are a pain aswell, I could square it here and there, to get rid of them, but I don't see where it's going again.

 

[Pere Callahan]

I only have time for the first one right now. Use a+b+c=1 ! and did you prove this inequality a+1/a+b+1/b+c+1/c >=10 ?

 

[Mark44]

For the first problem, pick some numbers and see how things work, and that might help you understand what is going on.

Given that a, b, and c > 0 and that a + b + c = 1, these numbers are necessarily less than 1.

One choice for all three variables is 1/3, from which the sum of the reciprocals is 9. If you choose a value for a that is larger than 1/3, then the other two variables will have to each be less than 1/3. For example, choose a = 1/2, b= 1/4, c = 1/4. What is the sum of the reciprocals? Is it larger or smaller than what you got before?

 

[ibc]

ya I get that, but I need to prove it, in a mathematical way... which I can't figure out how, if I take the case when a>1/3 then b or c could be >1/3 aswell, and then the other one must be smaller, of course the bigger I take the numbers, the smaller the third one will be, and thus the whole thing will be larger and larger, but I can't find a mathematical procedure to show it, I can't find the right connection between a,b,c and the fact that the sum of all equals 1 doesn't help me since I still have 1 inequality with 2 variables.

and I couldn't prove that a+1/a+b+1/b+c+1/c >=10 ?

 

[Pere Callahan]

The Attempt at a Solution

using the tip I could make it:
a+1/a+b+1/b+c+1/c ≥ 10
but that's as far as I got.

and I couldn't prove that a+1/a+b+1/b+c+1/c >=10 ?

You said you obtained that equation from the hint and than were stuck...

 

[Mark44]

I was able to prove it, although it took a bit of doing. The high points of my proof are as follows:
We want to find the minimum value of 1/a + 1/b + 1/c, subject to a, b, and c > 0 and a + b + c = 1.

Let f(x, y) = 1/x + 1/y + 1/(1 - x - y)
(The 3rd term uses the fact that the three numbers add up to 1.)

Take partials wrt x and y and set them to 0.
f_x = \frac{-1}{x^2} + \frac{1}{(1 - x - y)^2}
f_y = \frac{-1}{y^2} + \frac{1}{(1 - x - y)^2}

Setting both partials to 0 results eventually in y = \pm x. Since x and y must be positive, we take y = x, with both positive.

Now, f(x, y) = f(x, x) = g(x) = 1/x + 1/x + 1/(1 - 2x) = 2/x + 1/(1 - 2x)
= (3x - 2)/(x^2 - x)

g'(x) = -2(3x^2 -4x + 1) / (2x^2 - x)^2
= -2(3x - 1)(x - 1) / (2x^2 - x)

g'(x) = \frac{-2(3x^2 -4x + 1)}{(2x^2 - x)^2}
= \frac{-2(3x - 1)(x - 1)}{(2x^2 - x)}

From the above we see that g'(x) = 0 when x = 1/3 or when x = 1.
The graph of g has a local minimum when x = 1/3 and a local maximum when x = 1. The latter value is of no interest in this problem, since it implies that y = 1 and 1 - x - y = -1.

Putting everything together, we see that x = 1/3 and y = 1/3 (from previous work), so 1 - x - y = 1/3 as well.

Going back to the a, b, and c of the original problem, the value of 1/a + 1/b + 1/c is smallest when a = 1/3, b = 1/3, and c = 1/3.

What is that smallest value?
1/a + 1/b + 1/c = 3 + 3 + 3 = 9.
For any other values of a, b, and c that are positive and sum to 1,
1/a + 1/b + 1/c will be larger than 9, which shows that under the conditions of this problem,
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq 9
as was required.

I didn't use the suggestion that x + 1/x >= 2 for all x not equal to 0, so it's possible that a proof that uses this fact is simpler.

Mark

 

[ibc]

Let f(x, y) = 1/x + 1/y + 1/(1 - x - y)
(The 3rd term uses the fact that the three numbers add up to 1.)

Take partials wrt x and y and set them to 0.
f_x = \frac{-1}{x^2} + \frac{1}{(1 - x - y)^2}
f_y = \frac{-1}{y^2} + \frac{1}{(1 - x - y)^2}

Setting both partials to 0 results eventually in y = \pm x. Since x and y must be positive, we take y = x, with both positive.

Now, f(x, y) = f(x, x) = g(x) = 1/x + 1/x + 1/(1 - 2x) = 2/x + 1/(1 - 2x)
= (3x - 2)/(x^2 - x)



Mark

Thanks
though I didn't get that part,
what does the partial derivative of this equation means, and what does it mean that you set them to zero?

 

[Mark44]

If y = f(x) and f is differentiable, you can talk about dy/dx ( = f'(x)), the derivative of y with respect to x. Here f is a function of one variable, x.

If z = g(x, y), g is a function of two variables, so things get a bit more complicated. There is no longer the concept of "the derivative", but you can find partial derivatives with respect to x or with respect to y. When you take the partial of a function with respect to one of its variables, you treat the other variable as if it were a constant.

 

[weejee]

Use Cauchy-Schwartz inequality.

http://en.wikipedia.org/wiki/Cauchy–Schwarz_inequality

(a+b+c)(1/a+1/b+1/c)>=(sqrt(a)*sqrt(1/a)+sqrt(b)*sqrt(1/b)+sqrt(c)*sqrt(1/c))^2

 

[ibc]

If y = f(x) and f is differentiable, you can talk about dy/dx ( = f'(x)), the derivative of y with respect to x. Here f is a function of one variable, x.

If z = g(x, y), g is a function of two variables, so things get a bit more complicated. There is no longer the concept of "the derivative", but you can find partial derivatives with respect to x or with respect to y. When you take the partial of a function with respect to one of its variables, you treat the other variable as if it were a constant.

ya, but why do we set them both to be zero?
when partial derivative for y = partial derivative for x = 0, then? it's the function's minimum?
if so, then by saying x=y is enough to know that x=y=z=1/3

 

[Mark44]

If both partials are zero at a given point, you can't tell if you're at a local minimum, a local maximum, or neither. The function I was looking at, f(x, y) = 1/x + 1/y + 1/(1 - x - y) is such that I was reasonably certain that there was a minimum at (1/3, 1/3).

Just knowing that x = y wasn't enough to know that x, y, and z all had to be 1/3. There was some additional work that I did to establish that, which I described in my earlier post.

 

[ibc]

If both partials are zero at a given point, you can't tell if you're at a local minimum, a local maximum, or neither. The function I was looking at, f(x, y) = 1/x + 1/y + 1/(1 - x - y) is such that I was reasonably certain that there was a minimum at (1/3, 1/3).

Just knowing that x = y wasn't enough to know that x, y, and z all had to be 1/3. There was some additional work that I did to establish that, which I described in my earlier post.

but how can you assume there is a minimum at 1/3, 1/3? that's what we're trying to prove...

 

[Pere Callahan]

One could always check the Hessian to make sure that the critical point at (1/3,1/3) is truly a local minimum.

 

[ibc]

now I have no idea of what you are talking about
I think there should be a simpler solution, it's just the beginning of calculus course