Question

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.

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

are a,b,c,d positive integers?

Answer 5

yep!! sorry for non specifying that.

Answer 6

What about the n's ?

Answer 7

n is a set of natural numbers.

Answer 8

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?

Answer 9

sorry, I forgot to put +

Answer 10

That is the definition but we must prove it.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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.

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

I have to go too

Answer 19

good luck

Answer 20

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

Answer 21

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

Answer 22

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 \(a_n\).

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

@experimentX
Tell me your thoughts.

Answer 27

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.

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

@wio there is 'n' inside and outside ... its type x^x ... so you cannot use chain rule.

most likely you should be doing. I got that result from mathematica. Check again. More or less ... i had enough of this.