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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.

Mathematics
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Induction is the best way but I have no idea how the heck I prove it's decreasing.
If an+1 is < an, the the function is decreasing.
I know the definition ... how to show it? expanding using multinomial theorem is freaking ugly, not sure if induction works so easily.

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are a,b,c,d positive integers?
yep!! sorry for non specifying that.
What about the n's ?
n is a set of natural numbers.
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?
sorry, I forgot to put +
That is the definition but we must prove it.
nope!! that is always true ... there is ^n to consider of.
But substituting any real number should yield a decreasing pattern regardless.
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
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.
and a^n
I still think we have to use induction though. Seems tough though. Even the trivial case.
sorry ... have to go due to technical reasons. I'll be back in 6 hrs.
I have to go too
good luck
I think induction might work ... but it seems too ugly to work out.
if anyone found any trick to it .. please update it.
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 \(a_n\).
This is because we know \(\sqrt[n]{a}\ln(a) \) is going to be positive if \(a\) and \(n\) are positive.
Thus the right term will always be negative, the left term will always be positive, and when multiplied together we get a negative number.
I mean it seems a bit easier to notice that the derivative is always negative than to notice that the function is always decreasing.
@experimentX Tell me your thoughts.
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.
since\(0<1\leq a\), mutliplying by \(a^n\) yields \[\large a^n\leq a^{n+1}\\\large (a^n)^\frac{1}{n(n+1)}<(a^{n+1})^\frac{1}{n(n+1)}\]\[\Large 0<\sqrt[(n+1)]{a}<\sqrt[n]{a}\]
\[\large a_{n+1}=\sqrt[(n+1)] a +\sqrt[(n+1)] b +\sqrt[(n+1)] c +\sqrt[(n+1)] d <\sqrt[n] a +\sqrt[n] b +\sqrt[n] c +\sqrt[n] d=a_n\]
@experimentX Weird, I used the chain rule to get my derivative and I'm not sure where I went wrong.
Actually, I'm not so sure that I am wrong.
@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.
Ah I see, should be doing logarithmic differentiation.
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
If a=b=c=d =k then, a_n=k
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 \( a_{n+1} < a_n \)

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