Beautiful But So Dangerous

Algebra Level 4

x 3 y 4 z 3 ( x 4 + y 4 ) ( x y + z 2 ) 3 + y 3 z 4 x 3 ( y 4 + z 4 ) ( y z + x 2 ) 3 + z 3 x 4 y 3 ( z 4 + x 4 ) ( z x + y 2 ) 3 \small \frac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\frac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\frac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}

Over all positive reals x , y x, y and z z , the maximum of the expression above can be expressed as m n \dfrac{m}{n} , where m m and n n are coprime positive integers. Find m + n m+n .

Source: VMO 2014 problem 2 day 2.


The answer is 19.

Son Nguyen by Son Nguyen , 20, Vietnam

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1 solution

Munem Shahriar
Oct 1, 2017

Suppose M = x 3 y 4 z 3 ( x 4 + y 4 ) ( x y + z 2 ) 3 + y 3 z 4 x 3 ( y 4 + z 4 ) ( y z + x 2 ) 3 + z 3 x 4 y 3 ( z 4 + x 4 ) ( z x + y 2 ) 3 M = \dfrac{x^3y^4z^3}{(x^4+y^4)(xy+z^2)^3}+\dfrac{y^3z^4x^3}{(y^4+z^4)(yz+x^2)^3}+\dfrac{z^3x^4y^3}{(z^4+x^4)(zx+y^2)^3}

Put x y = a ; y z = b ; z x = c ; a b c = 1 \dfrac{x}{y} = a ; \dfrac{y}{z} =b; \dfrac{z}{x} = c; abc = 1

M = ∑ 1 ( a 4 + 1 ) ( b + c ) 3 M = \sum \dfrac{1}{(a^4 + 1)(b+c)^3}

We have

1 ( a 4 + 1 ) ( b + c ) 3 ≤ 1 ( a 2 + 1 ) 2 2 ( b + c ) 3 ≤ 1 ( a + 1 ) 4 8 ( b + c ) 3 = 8 ( a 4 + 1 ) ( b + c ) 3 ≤ 8 16 a 2 4 b c ( b 2 + c 2 ) \dfrac{1}{(a^4 + 1)(b+c)^3} \leq \dfrac{1}{\frac{(a^{2}+1)^{2}}{2}(b+c)^{3}} \leq \dfrac{1}{\frac{(a+1)^4}{8} (b+c)^3} = \dfrac{8}{(a^4 + 1)(b+c)^3} \leq \dfrac{8}{16a^2 4bc(b^2+c^2)}

= 1 8 a ( b 2 + c 2 ) ≤ 1 16 = \dfrac{1}{8a(b^2+c^2)} \le \dfrac{1}{16}

The maximum is 3 16 \dfrac{3}{16}

Hence 3 + 16 = 19 3 + 16 = \boxed{19}

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