Inspired by "Sparky" Kesa

Algebra Level 5

Let x , y , z x,y,z be reals such that x + y = k ( y + z ) > 0 x+y=k(y+z)>0 and z + x < 0 z+x<0 . Find the minimum value of t t such that for all k ≥ t k \ge t , there always exist x , y , z x, y, z satisfying x y + z + y x + z + z x + y = 1 \dfrac{x}{y+z} +\dfrac{y}{x+z} +\dfrac{z}{x+y}=1 . If min ( t ) = a + b c + d + e f g \text{min}(t)=\dfrac{\sqrt{a+b\sqrt{c}}+d+e\sqrt{f}}{g} (min t t is fully simplified) for positive integers a , b , c , d , e , f , g , a, b, c, d, e, f, g, submit your answer as the product a b c d e f g . abcdefg.

Inspiration.

P/S: Thanks to James Wilson for the conversation.


The answer is 965888.

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Steven Jim by Steven Jim , 17, Vietnam

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1 solution

Steven Jim
Nov 3, 2017

x y + z + y x + z + z x + y = 1 = > ( x + y + z ) ( 1 x + y + 1 y + z + 1 z + x ) = 4 \frac { x }{ y+z } +\frac { y }{ x+z } +\frac { z }{ x+y } =1=>(x+y+z)(\frac { 1 }{ x+y } +\frac { 1 }{ y+z } +\frac { 1 }{ z+x } )=4

Let x + y = a , y + z = b , z + x = c = > ( a + b + c ) ( 1 a + 1 b + 1 c ) = 8 x+y=a,y+z=b,z+x=c=>(a+b+c)(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } )=8 (1)

Note that if the set x 0 , y 0 , z 0 { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } satisfies the original equation, so does k x 0 , k y 0 , k z 0 k{ x }_{ 0 },k{ y }_{ 0 },k{ z }_{ 0 } . So WLOG, assume y + z = 1 y+z=1 .

Equation (1) becomes ( k + 1 + c ) ( k + 1 k + 1 c ) = 8 (k+1+c)(\frac { k+1 }{ k } +\frac { 1 }{ c } )=8

= > ( k + 1 + c ) ( c ( k + 1 ) + k ) = 8 c k =>(k+1+c)(c(k+1)+k)=8ck

= > c 2 ( k + 1 ) + c ( k 2 − 5 k + 1 ) + k ( k + 1 ) = 0 =>{ c }^{ 2 }(k+1)+c({ k }^{ 2 }-5k+1)+k(k+1)=0

= > ( c k + 1 ) 2 + 2 c k + 1 k 2 − 5 k + 1 2 k + 1 + ( k 2 − 5 k + 1 ) 2 4 ( k + 1 ) = k 4 − 14 k 3 + 19 k 2 − 14 k + 1 4 ( k + 1 ) =>(c\sqrt { k+1 } )^{ 2 }+2c\sqrt { k+1 } \frac { { k }^{ 2 }-5k+1 }{ 2\sqrt { k+1 } } +\frac { { (k }^{ 2 }-5k+1)^{ 2 } }{ 4(k+1) } =\frac { { k }^{ 4 }-14{ k }^{ 3 }+19{ k }^{ 2 }-14k+1 }{ 4(k+1) }

= > ( c k + 1 + k 2 − 5 k + 1 2 k + 1 ) 2 = k 4 − 14 k 3 + 19 k 2 − 14 k + 1 4 ( k + 1 ) =>(c\sqrt { k+1 } +\frac { k^{ 2 }-5k+1 }{ 2\sqrt { k+1 } } )^{ 2 }=\frac { { k }^{ 4 }-14{ k }^{ 3 }+19{ k }^{ 2 }-14k+1 }{ 4(k+1) }

= > ( 2 c ( k + 1 ) + k 2 − 5 k + 1 ) 2 = k 4 − 14 k 3 + 19 k 2 − 14 k + 1 =>(2c(k+1)+{ k }^{ 2 }-5k+1)^{ 2 }={ k }^{ 4 }-14{ k }^{ 3 }+19{ k }^{ 2 }-14k+1 .

Note that for c c to actually exist as a real number, k 4 − 14 k 3 + 19 k 2 − 14 k + 1 ≥ 0 { k }^{ 4 }-14{ k }^{ 3 }+19{ k }^{ 2 }-14k+1 \ge 0 .

Also note that k > 1 k>1 , as k = 1 = > k 4 − 14 k 3 + 19 k 2 − 14 k + 1 = − 7 k=1=>{ k }^{ 4 }-14{ k }^{ 3 }+19{ k }^{ 2 }-14k+1 = -7 .

= > k 2 − 14 k + 19 − 14 k + 1 k 2 ≥ 0 =>{ k }^{ 2 }-14k+19-\frac { 14 }{ k } +\frac { 1 }{ { k }^{ 2 } } \ge 0

= > ( k + 1 k ) 2 − 14 ( k + 1 k ) + 17 ≥ 0 =>{ (k+\frac { 1 }{ k } ) }^{ 2 }-14(k+\frac { 1 }{ k } )+17\ge 0 .

Let k + 1 k = x k+\frac { 1 }{ k }=x .

= > x 2 − 14 x + 17 ≥ 0 =>{ x }^{ 2 }-14x+17\ge 0

= > ( x − 7 ) 2 ≥ 32 =>{ (x-7) }^{ 2 }\ge 32

= > x ≥ 7 + 4 2 =>x\ge 7+4\sqrt{2}

= > k + 1 k ≥ 7 + 4 2 =>k+\frac { 1 }{ k } \ge 7+4\sqrt { 2 }

= > 4 k 2 − 4 k ( 7 + 4 2 ) + 4 ≥ 0 =>4{ k }^{ 2 }-4k(7+4\sqrt { 2 } )+4\ge 0

= > ( 2 k − ( 7 + 4 2 ) ) 2 ≥ 77 + 56 2 =>{ (2k-(7+4\sqrt { 2 } )) }^{ 2 }\ge 77+56\sqrt { 2 }

= > k ≥ 77 + 56 2 + 7 + 4 2 2 =>k \ge \frac { \sqrt { 77+56\sqrt { 2 } } +7+4\sqrt { 2 } }{ 2 } .

So for all k ≥ 77 + 56 2 + 7 + 4 2 2 k \ge \frac { \sqrt { 77+56\sqrt { 2 } } +7+4\sqrt { 2 } }{ 2 } , there always exists x , y , z x,y,z satisfying the original equation.

= > a b c d e f g = 965888 =>abcdefg=965888 .

P/S: Yes, the letter t t wasn't used at all in my solution... I typed it out as I found out that k k can be negative and there still exists a solution (just negate them) but when k = 1 k=1 , there exists no real solution. This is also the main reason I had to claim that k > 1 k>1 to proceed.

3 Helpful 0 Interesting 1 Brilliant 0 Confused

Excellent solution! My reasoning on Sharky's problem was definitely flawed. I'll let you know if I can break apart exactly what went wrong.

James Wilson - 3 years, 7 months ago

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I believe it was the fact that I assumed all three partial derivatives were zero. I figured the meat (or all) of the solution set of x y + z + y x + z + z x + y = 1 \frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1 was a connected region. Recall that that equation implies x 2 y + z + y 2 x + z + z 2 x + y = 0 \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=0 . But this does not ensure all three partial derivatives of x 2 y + z + y 2 x + z + z 2 x + y \frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y} over the same region are zero. That was the flaw in my thinking. The set would have to be "thicker."

James Wilson - 3 years, 7 months ago

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Yeah, haven’t really thought of it :)

Steven Jim - 3 years, 7 months ago

Don't you want to say t > 1 t>1 instead of k > 1 k>1 ?

James Wilson - 3 years, 7 months ago

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Well... Same thing after all :)

Steven Jim - 3 years, 7 months ago

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