Proving Inequality: y^5+y^2-7y+5\geq 0 for y\geq 1

  • Thread starter evagelos
  • Start date
  • Tags
    Inequality
In summary, the conversation discusses how to prove that the function y^5+y^2-7y+5 is greater than or equal to 0 for all y greater than or equal to 1. It is suggested to use calculus and determine the increasing and decreasing points of the function. One approach is to factor the polynomial, with Mathematica giving (y-1)^2*g_3(y) as the result. The conversation also mentions looking at the extremal values of the polynomial g_3(y) in the given interval.
  • #1
evagelos
315
0

Homework Statement



Prove that: [tex]y^5+y^2-7y+5\geq 0[/tex] ,for all [tex]y\geq 1[/tex]

Homework Equations


The Attempt at a Solution



[tex]y^5\geq 1[/tex] and [tex]y^2\geq 1[/tex] => [tex]y^5+y^2\geq 2[/tex].

Also [tex]-7y+5\leq -2[/tex] , and then?
 
Physics news on Phys.org
  • #2
evagelos said:

Homework Statement



Prove that: [tex]y^5+y^2-7y+5\geq 0[/tex] ,for all [tex]y\geq 1[/tex]



Homework Equations





The Attempt at a Solution



[tex]y^5\geq 1[/tex] and [tex]y^2\geq 1[/tex] => [tex]y^5+y^2\geq 2[/tex].

Also [tex]-7y+5\leq -2[/tex] , and then?
If f(y) = y5 + y2 - 7y + 5, note that f(1) = 0.

Look at f'(y) to determine where f is increasing and decreasing for y >= 1.
 
  • #3
Mark44 said:
If f(y) = y5 + y2 - 7y + 5, note that f(1) = 0.

Look at f'(y) to determine where f is increasing and decreasing for y >= 1.

O.K

f'(y)=[tex]5y^4+2y-7\geq 0[/tex] for [tex]y\geq 1[/tex] .

But how can this effect our f(y) ??
 
  • #4
f'(y) gives the slope at a point (y, f(y)) on the graph of f. If f'(y) > 0, the graph of f is increasing. If f'(y) < 0, the graph of f is decreasing.

We know that f(1) = 0. Is the graph of f going up or down from there?

BTW, this seems to be a calculus problem, so it should have been posted in the Calculus & Beyond section, not the Precalculus section.
 
  • #5
Write f(y) in terms of x=y-1≥0.

ehild
 
  • #6
Mark44 said:
f'(y) gives the slope at a point (y, f(y)) on the graph of f. If f'(y) > 0, the graph of f is increasing. If f'(y) < 0, the graph of f is decreasing.

We know that f(1) = 0. Is the graph of f going up or down from there?

BTW, this seems to be a calculus problem, so it should have been posted in the Calculus & Beyond section, not the Precalculus section.


you can have f(y)<0 and f'(y)>=0 ,so we cannot get a contradiction
 
  • #7
If the graph is always increasing, and its value at the leftmost point (y=1) is 0, then can it be negative to the right (y>1)? You can also follow ehild's suggestion to get a more direct answer.
 
  • #8
ehild said:
Write f(y) in terms of x=y-1≥0.

ehild

Like this?

[tex](x+1)^5+(x+1)^2-7(x+1)+5[/tex]
 
  • #9
Expand.
 
  • #10
evagelos said:
Like this?

[tex](x+1)^5+(x+1)^2-7(x+1)+5[/tex]

This is really the long way around. You have the derivative -- f'(y) = 5y4 + 2y - y. It's a very simple matter to show that f'(y) >= 0 for y >= 1, hence the graph is increasing for y >= 1, and you're pretty much done.
 
  • #11
Factor the polynomial. Mathemtica gives:

[tex]
y^{5} + y^{2} -7y+5 = (y - 1)^{2} \, g_{3}(y)
[/tex]

where [itex]g_{3}(y)[/itex] is a polynomial of 3rd digree. What are the extremal values of the polynomial [itex]g_{3}(y)[/itex] in the interval [itex]y \ge 1[/itex]?
 
  • #12
Mark44 said:
This is really the long way around.
It is not that difficult to expand, knowing the coefficients of (x+1)5 from Pascal's Triangle.

(x+1)5=x5+5x4+10x3+10x2+5x+1

ehild
 

1. How do you prove the inequality y^5+y^2-7y+5\geq 0 for y\geq 1?

To prove this inequality, we can use the method of mathematical induction. First, we can show that the statement holds true for the base case of y=1. Then, we can assume that the statement is true for some arbitrary value of y=k, and use this to prove that it is also true for y=k+1. This will prove the inequality for all values of y greater than or equal to 1.

2. What is the significance of the condition y\geq 1 in the inequality y^5+y^2-7y+5\geq 0?

The condition y\geq 1 is important because it restricts the domain of the inequality to values of y that are greater than or equal to 1. This allows us to use the method of mathematical induction to prove the inequality, as mentioned in the previous answer.

3. Can you use a counterexample to disprove the inequality y^5+y^2-7y+5\geq 0 for y\geq 1?

No, we cannot use a counterexample to disprove this inequality. The statement is true for the base case of y=1, and it is also true for all subsequent values of y due to the method of mathematical induction. Therefore, we cannot find a single counterexample that would disprove the inequality for all values of y\geq 1.

4. What is the role of y^5 in this inequality, and why is it necessary?

The term y^5 is necessary in this inequality because it helps to balance out the other terms and ensure that the statement holds true. Without this term, the inequality may not hold for all values of y\geq 1. It also contributes to the overall shape of the graph of the inequality, which can provide valuable insights into its behavior.

5. Are there any other methods that can be used to prove this inequality?

Yes, there are other methods that can be used to prove this inequality. For example, we can use the method of direct proof by manipulating the terms in the inequality to show that it holds true for all values of y\geq 1. We can also use the method of contradiction by assuming that the statement is false and then showing that this leads to a contradiction. However, the method of mathematical induction is often the most efficient and straightforward approach for proving inequalities like this one.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
2
Views
946
  • Precalculus Mathematics Homework Help
Replies
5
Views
762
  • Precalculus Mathematics Homework Help
Replies
4
Views
770
  • Precalculus Mathematics Homework Help
Replies
3
Views
714
  • Precalculus Mathematics Homework Help
Replies
7
Views
698
  • Precalculus Mathematics Homework Help
Replies
2
Views
848
  • Precalculus Mathematics Homework Help
Replies
1
Views
917
  • Precalculus Mathematics Homework Help
Replies
13
Views
1K
  • Precalculus Mathematics Homework Help
Replies
4
Views
2K
  • Precalculus Mathematics Homework Help
Replies
10
Views
860
Back
Top