<s>[INST] Disproving existence of real root in some interval for a quintic equation Disprove the statement: There is a real root of equation $\frac{1}{5}x^5+\frac{2}{3}x^3+2x=0$ on the interval (1,2).
I am not sure whether to prove by counter-example or by assuming the statement is true and then proving by contradiction. 
 [/INST] It's sufficient to show that:


*

*f(1) > 0

*For all n > 0: f(1 + n) > f(1) 
Step #2 means that the function is increasing at all points to the right of x=1.  From this you can say that f(x) doesn't equal zero (has a real root) in the range (1,2) because f(x) is always above 1 in that range.
Working out the algebra:
f(1 + n) = (1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5

Then to show that f(1 + n) > f(1), you have to show: 
(1/5 + 2/3 + 2) + 4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > (1/5 + 2/3 + 2)

Canceling the (1/5 + 2/3 + 2) gives:
4n + 4n^2 + (2 + 2/3)n^3 + n^4 + (1/5)n^5 > 0

It should be easy to see that as n increases, this inequality remains true, as long as n > 1.
Edit: here's one way to work this out:
Divide: by 4: n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0
We know that n>1 so we can add a second inequality which removes the initial n +:
n + n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > n^2 + (3/4)n^3 + (1/4)n^4 + (1/20)n^5 > 0

Subtract out the middle pieces to make the middle zero:
n > 0 > -n^2 - (3/4)n^3 - (1/4)n^4 - (1/20)n^5

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