## experimentX Group Title 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. one year ago one year ago

1. Dido525 Group Title

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

2. incomplte Group Title

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

3. experimentX Group Title

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

4. sparik1997 Group Title

are a,b,c,d positive integers?

5. experimentX Group Title

yep!! sorry for non specifying that.

6. Dido525 Group Title

7. experimentX Group Title

n is a set of natural numbers.

8. sparik1997 Group Title

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. sparik1997 Group Title

sorry, I forgot to put +

10. Dido525 Group Title

That is the definition but we must prove it.

11. experimentX Group Title

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

12. Dido525 Group Title

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

13. sparik1997 Group Title

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 Group Title

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. sparik1997 Group Title

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

16. Dido525 Group Title

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

17. experimentX Group Title

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

18. sparik1997 Group Title

I have to go too

19. sparik1997 Group Title

good luck

20. experimentX Group Title

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

21. experimentX Group Title

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

22. wio Group Title

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. wio Group Title

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

24. wio Group Title

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. wio Group Title

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. wio Group Title

27. experimentX Group Title

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. sirm3d Group Title

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. sirm3d Group Title

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

30. wio Group Title

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

31. wio Group Title

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

32. experimentX Group Title

@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. wio Group Title

Ah I see, should be doing logarithmic differentiation.

34. sauravshakya Group Title

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. sauravshakya Group Title

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

36. experimentX Group Title

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 $$a_{n+1} < a_n$$