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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} &\normalsize \text{Let a,b,c and d be four integers such that}\hspace{.33em}\\~\\ &a+b+c+d=4m+1\hspace{.33em}\\~\\ &k=a^2+b^2+c^2+d^2 \hspace{.33em}\\~\\ &\normalsize \text{where m is a positive integer}\hspace{.33em}\\~\\ &\normalsize \text{which one of the following is necessarily true}\hspace{.33em}\\~\\ & a.)\ \normalsize \text{The minimum possible value of k is }\ 4m^22m+1 \hspace{.33em}\\~\\ & b.)\ \normalsize \text{The minimum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ & c.)\ \normalsize \text{The maximum possible value of k is }\ 4m^22m+1 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{The maximum possible value of k is }\ 4m^2+2m+1 \hspace{.33em}\\~\\ \end{align}}\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2minimum value of m is \(1\) so \(k_{min}=5\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2lol i mean \((a+b+c+d)_{min}=5\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2cuz \(m\) is a positive integer

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2yeah but m is given + integer

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2i was thinking to use this AMGM inequality \(\large \color{black}{\begin{align} \dfrac{a^{2}+b^{2}+c^{2}+d^{2}}{4}\geq (a^{2}b^{2}c^{2}d^{2})^{(1/4)} \hspace{.33em}\\~\\ \end{align}}\) but \(a^{2}b^{2}c^{2}d^{2}\) is not given.

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you can use mathematical induction if you are a little clever. (and you know which is the correct answer, which there is a way, with calculus, to eliminate 3 out of the 4 possible answers)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2i didnt understand , answer given is option (b.) , btw m not clever

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1you can use Lagrange Multipliers to deduce that the only possible answer is b... then you can use induction on \(m\) to show that \(4m^2+2m+1\) is indeed the smallest \(k\) can be.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2unlucky me, i didnt studied calculus yet

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2can a b c d repeat ?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.2Like a b c d can be literraly any integers?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2yes \(\Large \{a,b,c,d\}\in \mathbb{Z}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(k\) can grow infinitely, because you can choose them such that\[a=b \\ c+d=4m+1\]and if you increase \(a\), \(k\) grows infinitely

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So you just need to look for minimum value of \(k\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2yes i need to eliminate one option from 2

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Now as you said, you just need to eliminate. You can do that by letting \(m = 1\).

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(a+b+c+d=5\), if \(m=1\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Let's try to find the minimum for real numbers a, b, c, d. I think we should use the expansion of \((a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2(ab + bc + cd + ac + ad + bd )\). So\[(4m+1)^2 = k + 2\lambda\]\[k = (4m + 1)^2  2\lambda \]To maximise \(k\), we minimise \(\lambda\). Since we know that \(a + b + c + d = 4m + 1\), we can see that \(\lambda\) will be minimum for \(a, b, c, d = m + \frac{1}{4}\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2\[k = (4m+1)^2  2 \cdot 6(m+ 1/4)^2\]\[= (4m + 1)^2  (12m^2 + 6m + 3/4)\]\[= 4m^2 + 2m + 1/4\]So the minimum of \(k\) for all real numbers is \(4m^2 + 2m + 1/4\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2So it of course cannot hold the value \(4m^2  2m + 1\), thus the answer is \(4m^2 + 2m + 1\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Is there some way to use AMGMHM here?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2but how can it be \(m+\dfrac14\) as all variables are given as integers

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I started by solving this problem for real numbers as there is no clear way to do so for integers (maybe there is  @mukushla might know about it).

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2u maximized \(k\) and this is \(k_{max}\) right ? \(k_{max} = (4m+1)^2  2 \cdot 6(m+ 1/4)^2\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2At our level, I don't think there's a satisfactory solution. As a rule, maximum/minimum occurs when all numbers are close to each other (possibly even equation). Here, I believe it occurs when three variables are \(m\) and one of them is \(m+1\).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Yes! This is \(k_{max}\) for real numbers.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2but as mukushla pointed out the \(k_max\) can be up to infinity , i just need to work on 2 options for \(k_{min}\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2Sorry, I meant \(k_{min}\). That is the \(k_{min}\) for reals and it coincides with \(\lambda_{max}\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Nice! with the options given, one way to get a quick answer is by plugging in \(m=1\) and see which option makes sense

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2for m=1 option a> 3 and option b>7 ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2option a doesn't makes sense , so option b is correct . nice

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.2I literally said the same thing in the first reply... :
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