# Inequality problem

1. Jun 25, 2010

### evagelos

1. The problem statement, all variables and given/known data

Prove that: $$y^5+y^2-7y+5\geq 0$$ ,for all $$y\geq 1$$

2. Relevant equations

3. The attempt at a solution

$$y^5\geq 1$$ and $$y^2\geq 1$$ => $$y^5+y^2\geq 2$$.

Also $$-7y+5\leq -2$$ , and then?

2. Jun 25, 2010

### Staff: Mentor

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. Jun 25, 2010

### evagelos

O.K

f'(y)=$$5y^4+2y-7\geq 0$$ for $$y\geq 1$$ .

But how can this effect our f(y) ??

4. Jun 25, 2010

### Staff: Mentor

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. Jun 25, 2010

### ehild

Write f(y) in terms of x=y-1≥0.

ehild

6. Jun 25, 2010

### evagelos

you can have f(y)<0 and f'(y)>=0 ,so we cannot get a contradiction

7. Jun 25, 2010

### Tedjn

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. Jun 25, 2010

### evagelos

Like this?

$$(x+1)^5+(x+1)^2-7(x+1)+5$$

9. Jun 25, 2010

### Tedjn

Expand.

10. Jun 25, 2010

### Staff: Mentor

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. Jun 25, 2010

### Dickfore

Factor the polynomial. Mathemtica gives:

$$y^{5} + y^{2} -7y+5 = (y - 1)^{2} \, g_{3}(y)$$

where $g_{3}(y)$ is a polynomial of 3rd digree. What are the extremal values of the polynomial $g_{3}(y)$ in the interval $y \ge 1$?

12. Jun 25, 2010

### ehild

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