Cool Inequality 7-90 Followers Problem!

Algebra Level 5

∣ a b ( a 2 − b 2 ) + b c ( b 2 − c 2 ) + c a ( c 2 − a 2 ) ∣ ≤ M ( a 2 + b 2 + c 2 ) 2 \large{|ab\left( { a }^{ 2 }-{ b }^{ 2 } \right) +bc\left( { b }^{ 2 }-{ c }^{ 2 } \right) +ca\left( { c }^{ 2 }-{ a }^{ 2 } \right) |\le M{ \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \right) }^{ 2 }}

For all real numbers a , b , c a,b,c the above inequality is satisfied. If the smallest value of constant M M can be expressed as

A B C \large{\frac { A }{ B } \sqrt { C } }

for some co-prime integers A A and B B and square-free number C C . Find A + B + C A+B+C


The answer is 43.

Department 8 by Department 8 , 27, Israel

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2 solutions

Note that M^2 is the maximum of

( a b ( a 2 − b 2 ) + a c ( c 2 − a 2 ) + b c ( b 2 − c 2 ) ) 2 ( a 2 + b 2 + c 2 ) 4 \frac{\left(a b \left(a^2-b^2\right)+a c \left(c^2-a^2\right)+b c \left(b^2-c^2\right)\right)^2}{\left(a^2+b^2+c^2\right)^4}

Letting { x = a + b + c , y = a b + a c + b c , a b c } \{x=a+b+c,y=a b+a c+b c,a b c\} , you can check that the above expression equals

x 2 ( − 4 x 3 z + x 2 y 2 + 18 x y z − 4 y 3 − 27 z 2 ) ( x 2 − 2 y ) 4 \frac{x^2 \left(-4 x^3 z+x^2 y^2+18 x y z-4 y^3-27 z^2\right)}{\left(x^2-2 y\right)^4}

I found this with a lot of computation (Mathematica). This is obviously not the best way to do it. Does anybody know of a nice way to figure this out without so much computation? Anyway, note that the above expression is quadratic in z. So maximizing this quadratic,we want to maximize

4 x 2 ( x 2 − 3 y ) 3 27 ( x 2 − 2 y ) 4 \frac{4 x^2 \left(x^2-3 y\right)^3}{27 \left(x^2-2 y\right)^4}

Setting Z = x 2 − 2 y Z=x^2-2 y :

4 ( Z − y ) 3 ( 2 y + Z ) 27 Z 4 \frac{4 (Z-y)^3 (2 y+Z)}{27 Z^4}

This is a quartic in y, so using calculus, we find its maximum occurs at y = − Z 8 y=-\frac{Z}{8} . Plugging this in yields M 2 = 81 512 M^2=\frac{81}{512} . Hence,

M = 9 16 2 M=\frac{9}{16 \sqrt{2}}

(Backtracking, we can easily show that this maximum can be achieved.)

This solution was very computation-heavy, so I welcome a better solution of any suggestions.

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Shourya Pandey
Nov 9, 2015

IMO 2006, Problem 3( or so).

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Please give a link for the solution

Department 8 - 5 years, 7 months ago

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Hint: a method is Lagrange Multipliers with some tweaking.

Daniel Prime - 5 years, 7 months ago

You can search the solutions on the website " www.imo.org". You can also search in google.

Priyanshu Mishra - 5 years, 6 months ago

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I know I have the solution but can you dare to right it here?

Department 8 - 5 years, 6 months ago

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@Department 8 – Yes, sure. here is my solution.

We first consider the cubic polynomial

P ( t ) = t b ( t 2 − b 2 ) + b c ( b 2 − c 2 ) + c t ( c 2 − t 2 ) P(t)=tb({ t }^{ 2 } - { b }^{ 2 }) + bc({ b }^{ 2 } - { c }^{ 2 }) + ct({ c }^{ 2 } - { t }^{ 2 }) .

It is easy to check that P ( b ) = P ( c ) = P ( − b − c ) = 0 P(b) = P(c) = P(-b - c) = 0 , and therefore

P ( t ) = ( b − c ) ( t − b ) ( t − c ) ( t + b + c ) P(t) = (b - c)(t - b)(t - c)(t + b + c)

since the cubic coefficient is b − c b - c . The left-hand side of the inequality can therefore be written in the form

a b ( a 2 − b 2 ) + b c ( b 2 − c 2 ) + c a ( c 2 − a 2 ) ∣ = ∣ P ( a ) ∣ = ∣ ( b − c ) ( a − b ) ( a − c ) ( a + b + c ) ∣ ab({ a }^{ 2 } - { b }^{ 2 }) + bc({ b }^{ 2 } - { c }^{ 2 }) + ca({ c }^{ 2 } - { a }^{ 2 })| = |P(a)| = |(b - c)(a - b)(a - c)(a + b + c)| . . . ( 1 ) ...(1)

Note that this expression is symmetric, and we can therefore assume a ≤ b ≤ c a ≤ b ≤ c without loss of generality. With this assumption,

|(a - b)(b - c)| = (b - a)(c - b) ≤ { \left( \frac { (b - a) + (c - b) }{ 2 } \right) }^{ 2 } = \frac { { (c - a) }^{ 2 } }{ 4 } , . . . ( 2 ) ...(2)

with equality if and only if b − a = c − b b - a = c - b , i.e. 2 b = a + c 2b = a + c . Also

{ \left( \frac { (c - b) + (b - a) }{ 2 } \right) }^{ 2 } \le \frac { { (c - b) }^{ 2 }{ + (b - a) }^{ 2 } }{ 2 } ,

or equivalently,

3 ( c − a ) 2 ≤ 2 ⋅ [ ( b − a ) 2 + ( c − b ) 2 + ( c − a ) 2 ] 3{ (c - a) }^{ 2 } ≤ 2·[{ (b - a) }^{ 2 } + { (c - b) }^{ 2 } + (c - a)^{ 2 }] , . . . ( 3 ) ...(3)

again with equality only for 2 b = a + c 2b = a + c . From ( 2 ) (2) and ( 3 ) (3) we get

∣ ( b − c ) ( a − b ) ( a − c ) ( a + b + c ) ∣ |(b - c)(a - b)(a - c)(a + b + c)|

≤ 1 4 ⋅ ∣ ( c − a ) 3 ( a + b + c ) \le \frac { 1 }{ 4 } ·|(c-a)^{ 3 }(a+b+c) )

= 1 4 . ( c − a ) 6 ( a + b + c ) 2 = \frac { 1 }{ 4 } .\sqrt { (c-a)^{ 6 }(a+b+c)^{ 2 } }

≤ 2 2 . ( ( 2. [ ( b − a ) 2 + ( c − a ) 2 + ( c − b ) 2 ] 3 ) 3 . ( a + b + c ) 2 4 ) 2 \le \frac { \sqrt { 2 } }{ 2 } .{ \left( \sqrt [ 4 ]{ { \left( \frac { 2.[{ (b - a) }^{ 2 } + { (c - a) }^{ 2 } + { (c - b })^{ 2 }] }{ 3 } \right) }^{ 3 }.{ (a + b + c) }^{ 2 } } \right) }^{ 2 }

By the weighted AM-GM inequality this estimate continues as follows:

∣ ( b − c ) ( a − b ) ( a − c ) ( a + b + c ) ∣ |(b - c)(a - b)(a - c)(a + b + c)|

≤ 9 2 32 . ( a 2 + b 2 + c 2 ) 2 \le \frac { 9\sqrt { 2 } }{ 32 } .{ ({ a }^{ 2 } + { b }^{ 2 } + { c }^{ 2 }) }^{ 2 } .

And hence M = 9 2 32 M = \boxed{\frac { 9\sqrt { 2 } }{ 32 }} .

Priyanshu Mishra - 5 years, 6 months ago

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@Priyanshu Mishra – @Andrew Ellinor please convert this into a solution

Department 8 - 5 years, 6 months ago

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@Department 8 – Yes, I was also seeking to write it in a solution column but did not find. So I replied you with the solution.

Priyanshu Mishra - 5 years, 6 months ago

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