I used desmos to graph the three functions anx actually the equation holds for x>=0.6782 whitch is the lower bound value that has been found. Great video as always.
I know that this isn't totally relevant because we are accepting all positive real values for n, but I thought it was interesting to note that 9n+8 can never be a aquare number for any integer valuenof n. 0,1,4,7 are the only square numbers mod 9
@@TheArtOfBeingANerdLet 9n+8=s² for some natural number s. Now look at the equation modulo 3: 0+2=s² so s²=2 (mod 3) which is not true because all perfect squares are 1 modulo 3.
@@TheArtOfBeingANerdAnother way to prove is to use @graf_paper method. Look at all the possible remainders of s² when divided by 9 and you'll notice that 8 is not in that list. s can leave a remainder of (0,1,2...8) when divided by 9 so just use modular arithmetic to find the remainders of s², here are some examples: If s=7 (mod 9) then s²=49=4(mod 9) If s=8 (mod 9) then s²=64=1(mod 9)
Using the well-known inequalities between geometric, arithmetic and squared means for not equal a, b, c: cbrt(abc) < (a+b+c)/3 < sqrt((a^2+b^2+c^2)/3) /cbrt stands for cube root/ for sqrt(n-1), sqrt(n), sqrt(n+1), we get 3*(cbrt(sqrt((n^3-n)) < sqrt(n-1)+sqrt(n)+sqrt(n+1) < 3*sqrt((n-1+n+n+1)/3) = sqrt(9n). For n>=4, it's easy to see that sqrt(9n-1) < 3*(cbrt(sqrt((n^3-n)), or equivalently (9n-1)^3 < 3^6(n^3-n): it leads to the quadratic 0 < (9n)^2 - 28*9n + 1/3 = (9n-14)^2 - 14^2 + 1/3, clearly true for n>=4. So for n>=4, sqrt(9n-1) < sqrt(n-1)+sqrt(n)+sqrt(n+1), and the same is easy to check "manually" for n=2, 3. Replacing n by n+1, we get the two desired inequalities: sqrt(9n+8) < sqrt(n)+sqrt(n+1)+sqrt(n+2) < sqrt(9n+9) implying the identity in question. Q.E.D.
How was this problem come up with? The '9' coefficient makes since (1+1+1)^2 = 9, but the constant at the end seems tricky to pinpoint, and I assume there are other non-integer constants which still satisfy the inequality. Is it posssible to generalise?
I found this posed as a problem to solve - there's a link in the description. Not sure how it was discovered though. It should be possible to generalise to include more terms. For example, some trial and error with replacing the "+8" term leads to sqrt{16n + 23} < sqrt{n} + sqrt{n+1} + sqrt{n+2} + sqrt{n+3} < sqrt{16n + 24}, so a similar result holds for floor{sqrt{16n + 23}}.
Great problem! Heer's a question: 3 sqrt(n) < sqrt(n) + sqrt(n+1) + sqrt(n+2) is true for all natural numbers; and sqrt(9n+8) < 3 sqrt(n+1) for all natural numbers. So why can't you just compare floor(3 sqrt(n)) < = floor(3 sqrt(n+1)), which is true for all natural numbers; hence, the stated problem is true for all natural numbers.
