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Algebra Level 5

( a 2 − a b + b 2 ) ( b 2 − b c + c 2 ) ( c 2 − a c + a 2 ) ≤ m n \large (a^2-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)\leq\frac{m}{n} If a , b a,b and c c are non-negative reals satisfying a + b + c = 2 a+b+c=2 and m m and n n are coprime positive integers, find m + n m+n .


The answer is 499.

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P C by P C , 20, Vietnam

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1 solution

Son Nguyen
Feb 13, 2016

Call the expression above is P By applying AM-GM we have: P ≤ a 2 b 2 ( a 2 − a b + b 2 ) = 4 9 ( 3 2 a b + 3 2 a b + a 2 − a b + b 2 3 ) P\leq a^2b^2(a^2-ab+b^2)=\frac{4}{9}(\frac{\frac{3}{2}ab+\frac{3}{2}ab+a^2-ab+b^2}{3}) = 4 9 ( a + b ) 6 27 ≤ 4 ( a + b + c ) 6 243 = 256 243 = m n =\frac{4}{9}\frac{(a+b)^6}{27}\leq \frac{4(a+b+c)^6}{243}=\frac{256}{243} =\frac{m}{n} So m + n = 499 \LARGE m+n=499

Another way Without losing the generally,we may assume that c=min{a,b,c} a 2 − a c + c 2 ≤ ( a + c 2 ) 2 a^2-ac+c^2\leq (a+\frac{c}{2})^2 b 2 − b c + c 2 ≤ ( b + c 2 ) 2 b^2-bc+c^2\leq (b+\frac{c}{2})^2 a 2 − a b + b 2 ≤ ( a + c 2 ) 2 − ( a + c 2 ) ( b + c 2 ) + ( b + c 2 ) 2 a^2-ab+b^2\leq (a+\frac{c}{2})^2-(a+\frac{c}{2})(b+\frac{c}{2})+(b+\frac{c}{2})^2 Let x = a + c 2 ; y = b + c 2 x=a+\frac{c}{2};y=b+\frac{c}{2} x + y = a + b + c = 2 x+y=a+b+c=2 P ≤ x 2 y 2 ( x 2 − x y + y 2 ) = x 2 ( 2 − x ) 2 [ x 2 − x ( 2 − x ) + ( 2 − x ) 2 ] P\leq x^2y^2(x^2-xy+y^2)=x^2(2-x)^2[x^2-x(2-x)+(2-x)^2] = 3 x 6 − 12 x 5 + 36 x 4 − 40 x 3 + 16 x 2 ≤ 256 243 = m n =3x^6-12x^5+36x^4-40x^3+16x^2\leq \frac{256}{243}=\frac{m}{n} So m + n = 499 \LARGE m+n=499

8 Helpful 0 Interesting 0 Brilliant 0 Confused

I concur with @Gurīdo Cuong . Do you mind explaining the first part ?(How do you use AM-GM in such mind-boggling way).

= 36 x 6 − 21 x 5 + 57 x 4 − 72 x 3 + 36 x 2 ≤ 256 243 = m n =36x^6-21x^5+57x^4-72x^3+36x^2\leq \frac{256}{243}=\frac{m}{n}

Why??

Also,

P ≤ x 2 y 2 ( x 2 − x y + y 2 ) = x 2 ( 2 − x 2 ) [ x 2 − x ( 2 − x ) + ( 2 − x ) 2 ] P\leq x^2y^2(x^2-xy+y^2)=x^2(2-x^2)[x^2-x(2-x)+(2-x)^2]

I believe you meant

P ≤ x 2 y 2 ( x 2 − x y + y 2 ) = x 2 ( 2 − x ) 2 [ x 2 − x ( 2 − x ) + ( 2 − x ) 2 ] P\leq x^2y^2(x^2-xy+y^2)=x^2(2-x)^2[x^2-x(2-x)+(2-x)^2]

ZK LIn - 5 years, 4 months ago

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I have fixed my problem.Maybe I'm wrong while I am typing

Son Nguyen - 5 years, 4 months ago

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Expanding x 2 ( 2 − x ) 2 [ x 2 − x ( 2 − x ) + ( 2 − x ) 2 ] x^2(2-x)^2[x^2-x(2-x)+(2-x)^2] gives you 3 x 6 − 12 x 5 + 36 x 4 − 40 x 3 + 1 6 2 3x^6-12x^5+36x^4-40x^3+16^2 .

P C - 5 years, 4 months ago

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@P C – After expanding everything and clearing terms,

a 2 − a b + b 2 ≤ ( a + c 2 ) 2 − ( a + c 2 ) ( b + c 2 ) + ( b + c 2 ) 2 a^2-ab+b^2\leq (a+\frac{c}{2})^2-(a+\frac{c}{2})(b+\frac{c}{2})+(b+\frac{c}{2})^2

a 2 − a b + b 2 ≤ a 2 + c 2 4 + c a − a b − a c 2 − b c 2 − c 2 4 + b 2 + c 2 4 + b c a^{2}-ab+b^{2} \leq a^{2}+\frac{c^{2}}{4}+ca-ab-\frac{ac}{2}-\frac{bc}{2}-\frac{c^{2}}{4}+b^{2}+\frac{c^{2}}{4}+bc

0 ≤ a c 2 + b c 2 + c 2 4 0\leq \frac{ac}{2}+\frac{bc}{2}+\frac{c^{2}}{4} happens to be correct.

However, the way the solution is phrased is easy to cause misunderstanding. Initially, I also struggled to see why the above inequality is true.

ZK LIn - 5 years, 4 months ago

can you further explain the AM-GM way please

P C - 5 years, 4 months ago

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Without losing the generally,we may assume that c=min{a,b,c} then using AM-GM

Son Nguyen - 5 years, 4 months ago

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Wow, you guys really love to solve inequalities problem. You should create a set that have all these questions!

By the way, @Gurīdo Cuong @MS HT @ZK LIn are you guys on Slack as well? It is a forum for most of the active users on this site, where all we do is talk about math and science. Plus, it's free as well! You should try it out.

Pi Han Goh - 5 years, 4 months ago

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@Pi Han Goh – yeap I will try

Son Nguyen - 5 years, 4 months ago

@Pi Han Goh – Sure, sounds like fun! I have quite a lot of free time now, so I will try.

ZK LIn - 5 years, 4 months ago

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