Christmas Problems: No.6

Algebra Level 3

We can observe that 5 a 2 + 5 b 2 + 5 c 2 + 5 d 2 + 5 e 2 = ( a + b + c + d + e ) 2 5{ a }^{ 2 }+5{ b }^{ 2 }+{ 5c }^{ 2 }+{ 5d }^{ 2 }+{ 5e }^{ 2 }=(a+b+c+d+e)^{ 2 } is always true when a = b = c = d = e a=b=c=d=e .

True or False?

There is a set of solutions ( a , b , c , d , e ) (a,b,c,d,e) which only contains distinct real numbers that satisfy the above equation.

False True

Kok Hao by Kok Hao , 18, Malaysia

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1 solution

Kok Hao
Dec 29, 2017

Relevant wiki: Titu's Lemma

5 a 2 + 5 b 2 + 5 c 2 + 5 d 2 + 5 e 2 = ( a + b + c + d + e ) 2 5{ a }^{ 2 }+5{ b }^{ 2 }+{ 5c }^{ 2 }+{ 5d }^{ 2 }+{ 5e }^{ 2 }=(a+b+c+d+e)^{ 2 }

( a 2 + b 2 + c 2 + d 2 + e 2 ) = ( a + b + c + d + e ) 2 5 { (a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }+{ d }^{ 2 }+{ e }^{ 2 })=\frac { (a+b+c+d+e)^{ 2 } }{ 5 }

( a 2 1 + b 2 1 + c 2 1 + d 2 1 + e 2 1 ) = ( a + b + c + d + e ) 2 1 + 1 + 1 + 1 + 1 (\frac { { a }^{ 2 } }{ 1 } +\frac { { b }^{ 2 } }{ 1 } +\frac { { c }^{ 2 } }{ 1 } +\frac { { d }^{ 2 } }{ 1 } +\frac { { e }^{ 2 } }{ 1 } )=\frac { (a+b+c+d+e)^{ 2 } }{ 1+1+1+1+1 }

This looks like Titu's lemma, which states that

a 1 b 1 + a 2 b 2 + . . . . . . + a n b n ≥ ( a 1 + a 2 + . . . . . . + a n ) 2 b 1 + b 2 + . . . . . . + b n \frac { { a }_{ 1 } }{ { b }_{ 1 } } +\frac { { a }_{ 2 } }{ { b }_{ 2 } } +......+\frac { { a }_{ n } }{ { b }_{ n } } \ge \frac { ({ a }_{ 1 }+{ a }_{ 2 }+......+{ a }_{ n })^{ 2 } }{ { b }_{ 1 }+{ b }_{ 2 }+......+{ b }_{ n } } , with equality happening at a 1 b 1 = a 2 b 2 = . . . . . . = a n b n \frac { { a }_{ 1 } }{ { b }_{ 1 } } =\frac { { a }_{ 2 } }{ { b }_{ 2 } } =......=\frac { { a }_{ n } }{ { b }_{ n } } .

Since all integers a a , b b , c c , d d and e e must be the same for equality to happen, there are no sets which have 5 distinct real numbers that satisfy the equation.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Note: You can do Cauchy Schwarz directly:

( 1 + 1 + 1 + 1 + 1 ) ( a 2 + b 2 + c 2 + d 2 + e 2 ) ≥ ( a + b + c + d + e ) 2 ( 1 + 1 + 1 + 1 + 1 ) ( a^2 + b^2 + c^2 + d^2 + e^2 ) \geq ( a + b + c + d + e) ^ 2

(Yes, I know that they are equivalent. I am talking about the "form" of the inequality.)

Calvin Lin Staff - 3 years, 5 months ago

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Thank you for the tip although I don't know how to apply this inequality in problems.

Kok Hao - 3 years, 5 months ago

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