## experimentX 3 years ago How to show that the the sequence $$\{ a_n \}$$ is a monotonically decreasing sequence. $a_n = { { \left (\frac { \sqrt [ n ]{ a } +\sqrt [ n ]{ b } +\sqrt [ n ]{ c } +\sqrt [ n ]{ d } }{ 4 } \right ) }^{ n } }$ It might not be monotonically decreasing sequence. I might be wrong in my assumption.

1. anonymous

Induction is the best way but I have no idea how the heck I prove it's decreasing.

2. anonymous

If an+1 is < an, the the function is decreasing.

3. experimentX

I know the definition ... how to show it? expanding using multinomial theorem is freaking ugly, not sure if induction works so easily.

4. anonymous

are a,b,c,d positive integers?

5. experimentX

yep!! sorry for non specifying that.

6. anonymous

7. experimentX

n is a set of natural numbers.

8. anonymous

do you agree that if wee show that $\sqrt[k]{a}+ \sqrt[k]{b} \sqrt[k]{c}+ \sqrt[k]{d}> \sqrt[k+1]{a}+ \sqrt[k+1]{b} \sqrt[k+1]{c}+ \sqrt[k+1]{d}$ then it's solved?

9. anonymous

sorry, I forgot to put +

10. anonymous

That is the definition but we must prove it.

11. experimentX

nope!! that is always true ... there is ^n to consider of.

12. anonymous

But substituting any real number should yield a decreasing pattern regardless.

13. anonymous

ok if we put n. we get some number in the brackets. and if we put n+1 , we get a smaller number in the brackets

14. experimentX

I am guessing quite opposite .. (nth root of a's)^(n) < (n+1 th root of s')^(n+1) .... most likely it's due to 4 down below.

15. anonymous

and a^n<b^n if a<b

16. anonymous

I still think we have to use induction though. Seems tough though. Even the trivial case.

17. experimentX

sorry ... have to go due to technical reasons. I'll be back in 6 hrs.

18. anonymous

I have to go too

19. anonymous

good luck

20. experimentX

I think induction might work ... but it seems too ugly to work out.

21. experimentX

if anyone found any trick to it .. please update it.

22. anonymous

Suppose: $$f(n) = a_n$$ Then: $f'(n) = (n-1)\left( \frac{\sqrt[n]{a}+\sqrt[n]{b}+\sqrt[n]{c}+\sqrt[n]{d}}{4} \right)^{n-1} \left( -\frac{\sqrt[n]{a}\ln(a) +\sqrt[n]{b}\ln(b)+\sqrt[n]{c}\ln(c)+\sqrt[n]{d}\ln(d)}{4n^2} \right)$When $$n > 1$$ we have $$f'(n)<0$$, thus $$f(n)$$ will always be decreasing, the same applied for $$a_n$$.

23. anonymous

This is because we know $$\sqrt[n]{a}\ln(a)$$ is going to be positive if $$a$$ and $$n$$ are positive.

24. anonymous

Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.

25. anonymous

I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.

26. anonymous

27. experimentX

seems legit ... but the derivative is $4^{-n} \left(a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right)^n \\ \left(-\text{Log}[4]-\frac{a^{\frac{1}{n}} \text{Log}[a]+b^{\frac{1}{n}} \text{Log}[b]+c^{\frac{1}{n}} \text{Log}[c]+d^{\frac{1}{n}} \text{Log}[d]}{\left(a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right) n}+\text{Log}\left[a^{\frac{1}{n}}+b^{\frac{1}{n}}+c^{\frac{1}{n}}+d^{\frac{1}{n}}\right]\right)$ now gotta show that this is always negative.

28. anonymous

since$$0<1\leq a$$, mutliplying by $$a^n$$ yields $\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}$$\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}$

29. anonymous

$\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n$

30. anonymous

@experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.

31. anonymous

Actually, I'm not so sure that I am wrong.

32. experimentX

@wio there is 'n' inside and outside ... its type x^x ... so you cannot use chain rule. |dw:1358029777544:dw| most likely you should be doing. I got that result from mathematica. Check again. More or less ... i had enough of this.

33. anonymous

Ah I see, should be doing logarithmic differentiation.

34. anonymous

when a=b=c=d=1 a_n = {(1+1+1+1)/4}^n a_n={4/4}^n a_n=1^n a_n=1 So, I dont think it is always decreasing

35. anonymous

If a=b=c=d =k then, a_n=k

36. experimentX

that's a special case ... where $$a_{n+1} = a_{n}$$ still you can call it monotonically decreasing ... because monotonic means $$a_{n+1} \le a_n$$ and strictly decreasing means $$a_{n+1} < a_n$$