Your answer is definitely not correct.
The answer may be 50, in my opinion because if you keep on increasing the value of variables \(a\) and \(e\) such that it goes arbitrarily near to 5 and accordingly decrease the other variables then the given sum will reach nearer and nearer to 50, but I dont think it will ever reach 50, or exceed 50.
Now, its coming back to the most confusing concept I have ever encountered that is the concept of Infinity.
–
Yatin Khanna
·
7 months, 4 weeks ago

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@Yatin Khanna
–
i agree with your reply..here i goes through the same confusing concept..
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Elite Limo
·
7 months ago

Since \(a,b,c,d,e\) need to be positive, there is no defined maximum value for \(I\). However, the supremum is \(50\). It is not possible for \(I\) to be greater than \(50\).
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Janardhanan Sivaramakrishnan
·
6 months, 3 weeks ago

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If doesn't matter which numbers we pick the maximum is 50.
If the numbers needs to be different the answer is 30
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Eugen Nica
·
7 months, 3 weeks ago

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@Eugen Nica
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If numbers are positive integer. If are not the answer is more complex.
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Eugen Nica
·
7 months, 3 weeks ago

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@Eugen Nica
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Firstly, 'positive numbers' refers to \(\mathbb{R}^+\) that is positive reals; thus the problem is not confined to positive integers. And if negative numbers are allowed then it is trivial that \(I\) can attain any positive value \(\geq 20 \frac{5}{6}\)
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Wen Z
·
7 months, 3 weeks ago

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@Wen Z silly me its how we say it in my language i meant quadratic arithmetic geometric and harmonic mean
–
Dragan Marković
·
7 months, 3 weeks ago

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@Dragan Marković
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if we're using those inequalities then I think that the question should be asking for the minimum value. If it is the maximum then the power means inequality does not obviously help as the maximising case in when the variables are equal and in this case the maximising case is not when the variables are equal.
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Wen Z
·
7 months, 3 weeks ago

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Our teacher gave us this task to solve by kagh inequalities. any ideas?
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Dragan Marković
·
7 months, 3 weeks ago

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@Dragan Marković
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whats that? A quick internet search yielded nothing
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Wen Z
·
7 months, 3 weeks ago

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(allowing for a variable to be \(0\))

Firstly prove that if

\(x+y=k\)

then to get the maximum value of \(x^2+y^2\) we have \(x,y=0,k\) in some order. The proof is quite simple; \(\begin{array} &x^2+y^2\\=&x^2+(k-x)^2\\=&2x(x-k)+k^2\end{array}\) where the first term is always non-positive and the second constant.

Now apply a few times (fix all variables except two) and get that the maximum of \(I\) is \(50\) if \(0\) is allowed. Since it is not allowed, just take the limit as all variables except two approach zero as the sum is continuous.
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Wen Z
·
7 months, 3 weeks ago

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I might be wrong but I think that the maximum value does not exist.
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Svatejas Shivakumar
·
7 months, 4 weeks ago

## Comments

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TopNewestYour answer is definitely not correct.

The answer may be 50, in my opinion because if you keep on increasing the value of variables \(a\) and \(e\) such that it goes arbitrarily near to 5 and accordingly decrease the other variables then the given sum will reach nearer and nearer to 50, but I dont think it will ever reach 50, or exceed 50.

Now, its coming back to the most confusing concept I have ever encountered that is the concept of Infinity. – Yatin Khanna · 7 months, 4 weeks ago

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– Elite Limo · 7 months ago

i agree with your reply..here i goes through the same confusing concept..Log in to reply

– Andre Tirta · 7 months, 4 weeks ago

Very good explanation YatinLog in to reply

Since \(a,b,c,d,e\) need to be positive, there is no defined maximum value for \(I\). However, the supremum is \(50\). It is not possible for \(I\) to be greater than \(50\). – Janardhanan Sivaramakrishnan · 6 months, 3 weeks ago

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If doesn't matter which numbers we pick the maximum is 50. If the numbers needs to be different the answer is 30 – Eugen Nica · 7 months, 3 weeks ago

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– Eugen Nica · 7 months, 3 weeks ago

If numbers are positive integer. If are not the answer is more complex.Log in to reply

– Wen Z · 7 months, 3 weeks ago

Firstly, 'positive numbers' refers to \(\mathbb{R}^+\) that is positive reals; thus the problem is not confined to positive integers. And if negative numbers are allowed then it is trivial that \(I\) can attain any positive value \(\geq 20 \frac{5}{6}\)Log in to reply

@Wen Z silly me its how we say it in my language i meant quadratic arithmetic geometric and harmonic mean – Dragan Marković · 7 months, 3 weeks ago

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– Wen Z · 7 months, 3 weeks ago

if we're using those inequalities then I think that the question should be asking for the minimum value. If it is the maximum then the power means inequality does not obviously help as the maximising case in when the variables are equal and in this case the maximising case is not when the variables are equal.Log in to reply

Our teacher gave us this task to solve by kagh inequalities. any ideas? – Dragan Marković · 7 months, 3 weeks ago

Log in to reply

– Wen Z · 7 months, 3 weeks ago

whats that? A quick internet search yielded nothingLog in to reply

(allowing for a variable to be \(0\))Firstly prove that if

\(x+y=k\)

then to get the maximum value of \(x^2+y^2\) we have \(x,y=0,k\) in some order. The proof is quite simple; \(\begin{array} &x^2+y^2\\=&x^2+(k-x)^2\\=&2x(x-k)+k^2\end{array}\) where the first term is always non-positive and the second constant.

Now apply a few times (fix all variables except two) and get that the maximum of \(I\) is \(50\) if \(0\) is allowed. Since it is not allowed, just take the limit as all variables except two approach zero as the sum is continuous. – Wen Z · 7 months, 3 weeks ago

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I might be wrong but I think that the maximum value does not exist. – Svatejas Shivakumar · 7 months, 4 weeks ago

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