Post on 26-Jan-2016
9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum Calvin Lin | Brilliant
https://brilliant.org/problems/findtuongsminimum/?group=3WveLY4nD4xV&ref_id=967118 1/5
Let , and be positive real numbers such that . What is the minimum value of
This problem is posed by Tuong N.
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Correct!You answered 64.
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From the HM-AM (harmonic mean-arithmetic mean) inequality, we can see that
But the LHS of the inequality is just , so we have . The
inequality holds if and only if , i.e. when . So the minimumof is indeed attainable, and our answer is .
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Apply the Cauchy-Schwarz's inequality . If ;
and , then the equality holds. So the mininum value of is 64.
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Find Tuong's minimum Algebra Level 4 165 points
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Derek Khu 1 year, 4 months ago
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Hưng Hồ 1 year, 4 months ago
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9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum Calvin Lin | Brilliant
https://brilliant.org/problems/findtuongsminimum/?group=3WveLY4nD4xV&ref_id=967118 2/5
From the AM-GM inequality, for any positive reals , note
with equality attained when . Now since ,
with equality attained if
With the added constraint that , we find
satisfies the conditions and achieves the minimum, which is .
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By Cauchy Inequality we have (1/x+1/y+4/z+16/t)*(x+y+z+t)>=(sqrt(1)+sqrt(1)+sqrt(4)+sqrt(16))^2=(1+1+2+4)^2=8^2=64
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Let the required expression be .
If we are able to find numbers, such that their sum is , i.e. 1; and the sum of theirreciprocals is , i.e. ; then using the inequality we can find the
minimum value of P. Here is the arithmetic mean of the numbers, and is the harmonicmean of the numbers.
This means that for each and , we have to find values of and such that is thecoefficient of and , and is the coefficient for . is the no. of times term is repeated.
The following equation is obtained for .
Hero P. 1 year, 4 months ago
10
Mihai Nipomici 1 year, 4 months ago
5
Pranshu Gaba 1 year, 4 months ago
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9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum Calvin Lin | Brilliant
https://brilliant.org/problems/findtuongsminimum/?group=3WveLY4nD4xV&ref_id=967118 3/5
and
and
Similarly, we get the following equations for and
and
and
and
So, we get the following terms,
.
The of these numbers is
The of these numbers is
Using the inequality, we get
Thus, we can conclude that the minimum value of , i.e. is 64.
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{(1/√x)^2+(1/√y)^2+(2/√z)^2+(4/√t)^2} {(√x)^2+(√y)^2+(√z)^2+(√t)^2}≥{(1/√x)(√x)+(1/√y)(√y)+(2/√z)(√z)+(4/√t)(√t)}^2 (1/x+1/y+4/z+16/t)(1)≥(1+1+2+4)^2(1/x+1/y+4/z+16/t)≥(8)^2 (1/x+1/y+4/z+16/t)(1)≥64. Done.Sorry i do not know how to use latexXD.
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By Cauchy-Schwarz, we have (x+y+z+t)(1/x+1/y+4/z+16/t)>=(1+1+2+4)^2=64. We know thatx+y+z+t=1, so we have 1/x+1/y+4/z+16/t>=64.
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Joshua Richard Theodoroes 1 year, 4 months ago
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Kevin Chang 1 year, 4 months ago
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Patrick Corn 1 year, 4 months ago
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9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum Calvin Lin | Brilliant
https://brilliant.org/problems/findtuongsminimum/?group=3WveLY4nD4xV&ref_id=967118 4/5
The minimum must be a local minimum on the hyperplane in 4-space. It's not onthe boundary of the region in question, because the quantity actually increases withoutbound as it approaches that boundary.
To find the local minima, we use Lagrange multipliers; they say that should be parallel to . Since all the variables are positive,
we get . This quickly leads to Since there is onlyone extremum in the region, it's either a global minimum or a global maximum, and it's clearly aglobal minimum because the expression blows up at the boundary. Plugging in, we get an answer of
.
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Considere a funcao f(x,y,z,t)=1/x + 1/y + 4/z + 16/t Existe uma restricao: x+y+z+t=1 Derivef(x,y,z,t) com relação a apenas uma das variaveis e iguale a uma constante L, multiplicada a derivadade g(x,y,z,t) com relação apenas a essa mesma variável Para x, temos: A derivada de f apenas emrelacao a x é a derivada de 1/x: -1/x² A derivada de g apenas em relação a x é a derivada de x: 1 Logovoce escreve: -1/x² = L1 Faca o mesmo para todas as variaveis, utilizando o mesmo L: -1/y² = L1-4/z² = L1 -16/t² = L1 Chame L de -1/c^2 -1/x² = L = -1/c² => x = c -1/y² = L = -1/c² => y= c -4/z²= L = -1/c² => z = 2c -16/t² = L = -1/c² => t = 4c Agora, use o fato de x+y+z+t = 1 pra achar c.c+c+2c+4c = 1 8c=1 c=1/8 Assim, temos: x=1/8 y=1/8 z=21/8=1/4 t=41/8=1/2 f(1/8,1/8,1/4,1/2) =8+8+44+162=64
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x<=y<z<t and we get x=1/8 y=1/8 z=1/4 t=1/2
1/x + 1/y + 4/z + 16/t = 8 + 8 + 16 + 32 = 64
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Once we have and , and x,y,z,t belong to , we may conclude
that:
I- If one of them belonged to , the sum would not be 1. So, they belong to .
II - Since they're fractions, they may be written as .
III- . And .
IV- To reach the lowest sum, must be the lowest possible. For this, must be the lowest. So,
.Then, .
V - So .
VI- .
Pedro Mourão 1 year, 4 months ago
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Syifa Adelia 1 year, 4 months ago
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Marciel Silva de Almeida 1 year, 4 months ago
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9/25/2015 Algebra Problem on Applying the Cauchy Schwarz Inequality: Find Tuong's minimum Calvin Lin | Brilliant
https://brilliant.org/problems/findtuongsminimum/?group=3WveLY4nD4xV&ref_id=967118 5/5
Using the same metod, we may infer, firstly, that would be . But, we have to consider that for , must be the lowest possible, and, in this case( ), .
Obviously, we may say that , because different numbers would make that would
make for every variable( ).
When we affirm that , . Then .
So, the minimum number is 64.
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