Minimal value

Algebra Level 5

Given an integer n ≥ 2 \displaystyle n \ge 2 .The minimal value of x 1 5 x 2 + x 3 + … + x n + x 2 5 x 1 + x 3 + … + x n + … + x n 5 x 1 + … x n − 1 \displaystyle \frac{x_{1}^{5}}{x_{2}+{x_{3}}+…+x_{n}} + \frac{x_{2}^{5}}{x_{1}+x_{3}+…+x_{n}} +…+ \frac{x_{n}^{5}}{x_{1}+…x_{n-1}} ,for positive real numbers x 1 , x 2 , … , x n x_{1},x_{2},…,x_{n} subject to the condition that sum of their squares is 1 , can be expressed as α n ( n − β ) \displaystyle \frac{\alpha}{n(n-\beta)} where α \displaystyle \alpha and β \displaystyle \beta are positive integers then what is the value of α + β \displaystyle \alpha+\beta


The answer is 2.0.

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Abhishek Singh by Abhishek Singh , 24, India

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4 solutions

Mathh Mathh
Jul 19, 2014

In this full proof I will use the fact that

x 1 x 2 + x 1 x 3 + ⋯ + x 1 x n + x 1 x 2 + x 2 x 3 + ⋯ + x 2 x n + ⋯ + x 1 x n + x 2 x n + ⋯ + x n − 1 x n \displaystyle x_1x_2+x_1x_3+\cdots +x_1x_n +x_1x_2+x_2x_3+\cdots+x_2x_n+\cdots+x_1x_n+x_2x_n+\cdots+x_{n-1}x_n = ( x 1 + x 2 + ⋯ + x n ) 2 − ( x 1 2 + x 2 2 + ⋯ + x n 2 ) \displaystyle = (x_1+x_2+\cdots+x_n)^2-(x_1^2+x_2^2+\cdots+x_n^2) = ( x 1 + x 2 + ⋯ + x n ) 2 − 1 \displaystyle =(x_1+x_2+\cdots+x_n)^2-1

Here's the proof:

x 1 5 x 2 + x 3 + ⋯ + x n + x 2 5 x 1 + x 3 + ⋯ + x n + ⋯ + x n 5 x 1 + x 2 + ⋯ + x n − 1 \displaystyle \frac{x_1^5}{x_2+x_3+\cdots +x_n}+\frac{x_2^5}{x_1+x_3+\cdots+x_n}+\cdots+\frac{x_n^5}{x_1+x_2+\cdots+x_{n-1}} = x 1 6 x 1 x 2 + x 1 x 3 + ⋯ + x 1 x n + x 2 6 x 1 x 2 + x 2 x 3 + ⋯ + x 2 x n + ⋯ + x n 6 x 1 x n + x 2 x n + ⋯ + x n − 1 x n \displaystyle = \frac{x_1^6}{x_1x_2+x_1x_3+\cdots +x_1x_n}+\frac{x_2^6}{x_1x_2+x_2x_3+\cdots+x_2x_n}+\cdots+\frac{x_n^6}{x_1x_n+x_2x_n+\cdots+x_{n-1}x_n} ≥ H o ¨ lder ( x 1 2 + x 2 2 + ⋯ + x n 2 ) 3 n ( x 1 x 2 + x 1 x 3 + ⋯ + x 1 x n + x 1 x 2 + x 2 x 3 + ⋯ + x 2 x n + ⋯ + x 1 x n + x 2 x n + ⋯ + x n − 1 x n ) \displaystyle \stackrel{\text{Hölder}}\ge \frac{(x_1^2+x_2^2+\cdots+x_n^2)^3}{n(x_1x_2+x_1x_3+\cdots +x_1x_n +x_1x_2+x_2x_3+\cdots+x_2x_n+\cdots+x_1x_n+x_2x_n+\cdots+x_{n-1}x_n)} = 1 n ( x 1 x 2 + x 1 x 3 + ⋯ + x 1 x n + x 1 x 2 + x 2 x 3 + ⋯ + x 2 x n + ⋯ + x 1 x n + x 2 x n + ⋯ + x n − 1 x n ) \displaystyle = \frac{1}{n(x_1x_2+x_1x_3+\cdots +x_1x_n +x_1x_2+x_2x_3+\cdots+x_2x_n+\cdots+x_1x_n+x_2x_n+\cdots+x_{n-1}x_n)} = 1 n ( ( x 1 + x 2 + ⋯ + x n ) 2 − 1 ) \displaystyle =\frac{1}{n((x_1+x_2+\cdots+x_n)^2-1)} = 1 n ( x 1 + x 2 + ⋯ + x n ) 2 − n \displaystyle =\frac{1}{n(x_1+x_2+\cdots+x_n)^2-n} ≥ Cauchy-Schwarz 1 n 2 ( x 1 2 + x 2 2 + ⋯ + x n 2 ) − n \displaystyle \stackrel{\text{Cauchy-Schwarz}}\ge \frac{1}{n^2(x_1^2+x_2^2+\cdots+x_n^2)-n} = 1 n ( n − 1 ) \displaystyle =\frac{1}{n(n-1)}

Since equality can be reached when x 1 = x 2 = ⋯ = x n = n n x_1=x_2=\cdots=x_n=\frac{\sqrt{n}}{n} , this is the lower bound of the original expression.

Which means α = β = 1 ⟹ α + β = 2 \alpha=\beta=1\implies \alpha+\beta =\boxed{2} .

6 Helpful 0 Interesting 1 Brilliant 0 Confused

thats the way i did it

huge wolverine - 6 years, 10 months ago

There's lots of ways to formulate a proof for this... I basically introduced S = sum of all variables and an ordering for reasons of symmetry, applied Chebyshev, then AM-HM (that's how the n-1 factor popped up) and then Chebyshev again twice, introducing the constraint in my very last step. This made it easy to generalize the problem to a constraint like 'sum of k-th powers equal U' and instead of 5, any power P in each numerator for which P - 1 = 0 mod k.

T B - 6 years, 10 months ago
Aaaaa Bbbbb
Jul 18, 2014

Use the case of n=2 and n=3 in combination with the expression: α n × ( n − β ) \frac{\alpha}{n \times (n-\beta)} It is easy to get: α = β = 1 ⇒ α + β = 2 \alpha=\beta=1 \Rightarrow \alpha + \beta = \boxed{2}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

I wanted to add another solution, but I can't see how. I was using the android app. Maybe @Calvin Lin or @Peter Taylor could help me.

Solution: Define S = x 1 + x 1 + ⋯ + x n S= x_1 + x_1 + \cdots + x_n and F F to be the expression in the problem. By the Cauchy-Schwarz inequality, we get:

( ∑ i = 1 n x i 5 S − x i ) ( ∑ i = 1 n ( S − x i ) ) ≥ ( ∑ x i 2.5 ) \displaystyle ( \sum_{i=1}^n \frac{x_i^5}{S-x_i} ) ( \sum_{i=1}^n (S-x_i)) \geq ( \sum x_i^{2.5} )

⇒ F ( n − 1 ) S ≥ ( ∑ x i 2.5 ) \Rightarrow F (n-1) S \geq ( \sum x_i^{2.5} )

⇒ F ≥ ( ∑ x i 2.5 ) ( n − 1 ) S \Rightarrow F \geq \frac {( \sum x_i^{2.5} )}{(n-1) S} .

Now using Chebyshev's inequality we get:

⇒ ( ∑ x i 2.5 ) S ≥ ∑ x i 2 n = 1 n \Rightarrow \frac {( \sum x_i^{2.5} )}{ S} \geq \frac{ \sum x_i^2}{n} = \frac {1}{n} .

Using these we get the result α + β = 2. □ \alpha + \beta = 2. \square

A Brilliant Member - 6 years, 11 months ago

u can take the case n=2 and then assign cos x and sin x as the two values . taking x as 45 degrees and applying it in the equation you get a value which on inspection would lead to the answer

huge wolverine - 6 years, 10 months ago
Nikola Djuric
Jun 19, 2015

Notist that (x1+x2+..x(n-1)) / (n-1) <√( (x1²+x2²+..x(n-1)²) / (n-1) )
so x1+x2+...+x(n-1) <√(n-1) √(1-x(n)²) A-H inequality for others also, so whole expression is >= 1/√(n-1) * sum(Xi^5)/√(1-Xi²) previous equality is achieved when x1=x2=...=x(n)=x so from sum(xi²)=1 we get nx²=1,x=1/√n,so =1/√(n-1) *n (1/√n^5)/√(1-1/n)=1/n(n-1) so a=1,b=1,a+b=2

1 Helpful 0 Interesting 0 Brilliant 0 Confused
Rahul Saxena
Mar 26, 2015

I myself used the case of n=2 and n=3 and got the correct answer.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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