Zi Song's iterated polynomial

Algebra Level 5

For all ordered triples ( p , q , r ) (p,q,r) , define the polynomial f p , q , r ( x ) = x 3 − p x 2 + q x − r . f_{p, q, r}(x) = x^{3} - px^{2} + qx - r. Let a 1 , a 2 , a 3 , b 1 , b 2 , b 3 , c 1 , c 2 , c 3 a_1, a_2, a_3, b_1, b_2, b_3, c_1, c_2, c_3 be (not necessary distinct) positive real numbers such that the roots of f a 1 , a 2 , a 3 ( x ) f_{a_1, a_2, a_3}(x) are b 1 , b 2 , b 3 b_1, b_2, b_3 and the roots of f b 1 , b 2 , b 3 ( x ) f_{b_1, b_2, b_3}(x) are c 1 , c 2 , c 3 c_1, c_2, c_3 . What is the maximum possible value of 9 b 3 3 b 1 + 3 + 4 + 3 b 1 + 2 b 2 + b 3 a 1 + 1 ? \frac{9\sqrt[3]{b_3}}{b_1 + 3} + \frac{4+3b_1 + 2b_2 + b_3}{a_1 + 1}?

This problem is posed by Zi Song Y .


The answer is 4.

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Calvin Lin by Calvin Lin

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2 solutions

Zi Song Yeoh
Sep 22, 2013

Let x = c 1 , y = c 2 . z = c 3 , a = b 1 , b = b 2 , c = b 3 , d = a 1 x = c_1, y = c_2. z = c_3, a = b_1, b = b_2, c = b_3, d = a_1 . By Vieta's Formula, x + y + z = a , x y + y z + z x = b , x y z = c , a + b + c = d x + y + z = a, xy + yz + zx = b, xyz = c, a + b + c = d . Hence, x x + 1 + y y + 1 + z z + 1 = 3 − ( 1 x + 1 + 1 y + 1 + 1 z + 1 ) = 3 − ( x y + y z + x z ) + 2 ( x + y + z ) + 3 x y z + ( x y + y z + x z ) + ( x + y + z ) + 1 = 3 − 2 a + b + 3 a + b + c + 1 = 3 − 2 a + b + 3 d + 1 \frac{x}{x + 1} + \frac{y}{y + 1} + \frac{z}{z + 1} = 3 - (\frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1}) = 3 - \frac{(xy + yz + xz) + 2(x + y + z) + 3}{xyz + (xy + yz + xz) + (x + y + z) + 1} = 3 - \frac{2a + b + 3}{a + b + c + 1} = 3 - \frac{2a + b + 3}{d + 1} . Thus, x x + 1 + y y + 1 + z z + 1 = 3 − 2 a + b + 3 d + 1 = 4 − 4 + 3 a + 2 b + c d + 1 \frac{x}{x + 1} + \frac{y}{y + 1} + \frac{z}{z + 1} = 3 - \frac{2a + b + 3}{d + 1} = 4 - \frac{4 + 3a + 2b + c}{d + 1} .

Thus, by the Cauchy-Schwarz Inequality, we get

4 − 4 + 3 a + 2 b + c d + 1 = ( x ) 2 x + 1 + ( y ) 2 y + 1 + ( z ) 2 z + 1 ≥ ( x + y + z ) 2 x + y + z + 3 ≥ ( 3 x y z 3 ) 2 a + 3 = 9 c 3 a + 3 4 - \frac{4 + 3a + 2b + c}{d + 1} = \frac{(\sqrt{x})^{2}}{x + 1} + \frac{(\sqrt{y})^{2}}{y + 1} + \frac{(\sqrt{z})^{2}}{z + 1} \ge \frac{(\sqrt{x} + \sqrt{y} + \sqrt{z})^{2}}{x + y + z + 3} \ge \frac{(3\sqrt[3]{\sqrt{xyz}})^2}{a + 3} = \frac{9\sqrt[3]{c}}{a + 3} , where the last inequality holds by AM-GM.

Thus, 9 c 3 a + 3 + 4 + 3 a + 2 b + c d + 1 = 9 b 3 3 b 1 + 3 + 4 + 3 b 1 + 2 b 2 + b 3 a 1 + 1 ≤ 4 \frac{9\sqrt[3]{c}}{a + 3} + \frac{4 + 3a + 2b + c}{d + 1} = \frac{9\sqrt[3]{b_3}}{b_1 + 3} + \frac{4 + 3b_1 + 2b_2 + b_3}{a_1 + 1}\le 4 . Equality holds when c 1 = c 2 = c 3 = 1 c_1 = c_2 = c_3 = 1 . (and possibly other values.) Thus, N = 4 N = 4 .

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Moderator note:

Nice solution, and a nice problem!

The first part of the solution looked like you knew the solution & accordingly framed the problem, & ah! Yes you did! :p

A Brilliant Member - 7 years, 8 months ago

the problem is somewhat not nice, since the way you add and subtract must be done after you know the answer is 4

Cuong Doan - 7 years, 8 months ago

ZI SONG I LOVE YOU!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Cody Johnson - 7 years, 5 months ago

wow nice long explanation god I wish I was as smart as people in this site, I just guessed it XD well not totally guessed it. I just got

b1 = c1+c2+c3

b2 = c1c2+c1c3+c2c3

b3 = c1c2c3

so I figured that since the denominator in the first half has b1 in it and the numerator has b3 in it, the result which maximizes the equation has to be the result which gets the biggest b3 while getting the smallest possible b1. if c1=c2=c3 it will get that result since b3/b1 will be (c^3)/3c which yields the biggest possible value. and then I just started doing c=2,c=3.c=4... and got 4 as the biggest result. "Cauchy-Schwarz Inequality"... I still have a long way to go I guess

Daniel Magen - 7 years, 8 months ago

Can you elaborate a little on how Cauchy-Schwarz was applied in the second-to-last line? You seem to have applied it to the numerators and just added the denominators.

Dan Rubery - 7 years, 8 months ago

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( a 1 2 b 1 + a 2 2 b 2 + . . . + a n 2 b n ) ( b 1 + b 2 + . . . + b n ) ≥ ( a 1 + a 2 + . . . + a n ) 2 (\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + ... + \frac{a_n^2}{b_n})(b_1 + b_2 + ... + b_n) \ge (a_1 + a_2 + ... + a_n)^2 , by the Cauchy-Schwarz inequality.

Zi Song Yeoh - 7 years, 8 months ago

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Ah, I see now, thank you.

Dan Rubery - 7 years, 8 months ago

Im sorry but i still couldnt get the cauchy-schwarz inequality part. It's |x.y| <= ||x|| ||y||

Edmund Heng - 7 years, 8 months ago

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@Edmund Heng – If you write out the components of x as x 1 , x 2 , . . . , x n x_1, x_2, ..., x_n and y as y 1 , y 2 , . . . , y n y_1, y_2,..., y_n , you can manipulate this into another common form of the inequality: ( Σ x i y i ) 2 ≤ ( Σ x i 2 ) ( Σ y i 2 ) (\Sigma x_i y_i)^2 \leq (\Sigma x_i^2) (\Sigma y_i^2) .

Using x i = a i b i x_i = \frac{a_i}{\sqrt{b_i}} and y i = b i y_i = \sqrt{b_i} gives the inequality above.

Dan Rubery - 7 years, 8 months ago

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@Dan Rubery – Thanks!! i got it! Nice one, never thought of this usage before

Edmund Heng - 7 years, 8 months ago

By Vieta's relations, we get: b 1 = ∑ c i ; b 2 = ∑ c i c j ; b 3 = c 1 c 2 c 3 b_1 = \sum c_i ; b_2 = \sum c_ic_j ; b_3 = c_1c_2c_3 & a 1 = ∑ b i a_1 = \sum b_i (that's all we need)

Observe that: a 1 + 1 = ( 1 + c 1 ) ( 1 + c 2 ) ( 1 + c 3 ) a_1+1 = (1+c_1)(1+c_2)(1+c_3) & that by A.M-G.M,

c 1 + c 2 + c 3 3 ≥ c 1 c 2 c 3 3 ⇒ 3 + c 1 + c 2 + c 3 ≥ 3 + 3 c 1 c 2 c 3 3 \frac{c_1+c_2+c_3}{3} \geq \sqrt[3]{c_1c_2c_3} \Rightarrow 3+c_1+c_2+c_3 \geq 3+3\sqrt[3]{c_1c_2c_3} with equality iff c 1 = c 2 = c 3 c_1=c_2=c_3

Let S : = 9 b 3 3 b 1 + 3 + 4 + 3 b 1 + 2 b 2 + b 3 a 1 + 1 S:= \frac{9 \sqrt[3]{b_3}}{b1+3} + \frac{4+3b_1+2b_2+b_3}{a_1+1}

Plugging in all values we get by rearrangement,

S = 9 c 1 c 2 c 3 3 c 1 + c 2 + c 3 + 3 + 4 + 3 ∑ c i + 2 ∑ c i c j + c 1 c 2 c 3 ( 1 + c 1 ) ( 1 + c 2 ) ( 1 + c 3 ) S= \frac{9 \sqrt[3]{c_1c_2c_3}}{c_1+c_2+c_3+3} + \frac{4+3\sum c_i+2\sum c_ic_j+c_1c_2c_3}{(1+c_1)(1+c_2)(1+c_3)}

≤ 3 c 1 c 2 c 3 3 1 + c 1 c 2 c 3 3 + 1 + 1 1 + c 1 + 1 1 + c 2 + 1 1 + c 3 \leq \frac{3 \sqrt[3]{c_1c_2c_3}}{1+\sqrt[3]{c_1c_2c_3}} + 1 +\frac{1}{1+c_1} +\frac{1}{1+c_2}+ \frac{1}{1+c_3}

= 4 + 1 1 + c 1 + 1 1 + c 2 + 1 1 + c 3 − 3 1 + c 1 c 2 c 3 3 = 4 +\frac{1}{1+c_1} +\frac{1}{1+c_2}+ \frac{1}{1+c_3} - \frac{3}{1+\sqrt[3]{c_1c_2c_3}}

For equality c 1 = c 2 = c 3 = c c_1=c_2=c_3=c ,

S m a x = 4 + 3 1 + c − 3 1 + c = 4 S_{max}= 4+ \frac{3}{1+c} - \frac{3}{1+c} = 4 .

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Unfortunately, for you argument to work, you must have 1 1 + c 1 + 1 1 + c 2 + 1 1 + c 3 − 3 1 + c 1 c 2 c 3 3 ≤ 0 \frac{1}{1+c_1} +\frac{1}{1+c_2}+ \frac{1}{1+c_3} - \frac{3}{1+\sqrt[3]{c_1c_2c_3}}\leq 0 for all c 1 , c 2 , c 3 c_1,c_2,c_3 (not necessarily equal).

And this is actually false, for example for c 1 = 8 , c_1=8, and c 2 = c 3 = 1 c_2=c_3=1 . What this really means is that you gave up too much room when you used the AM-GM inequality the way you did. This is a subtle inequality, with no known simple proof.

Alexander Borisov - 7 years, 8 months ago

The AM-GM inequality was correct, but I think you're pointing out at the fact that after using the inequality, I'm adding variables to both sides, which were involved in the inequality. I thought that since I'm adding the same value(whatever that might be) to both sides,there won't be a problem,though not expecting such a disparity. However the inequality in that case, as you have pointed, will be still correct,except the equality part.

You say S < 4 + S < 4 + some positive quantity, but S ≠ 4 + S \neq 4+ that quanitity, so basically you don't achieve an upper bound(with which S might be equal to). Please correct me if I'm missing something & also clarify how I 'gave up too much room'...

I think you mean ideally the sequence should approach it's maximum from negative side, but'<' doesn't really provide bounding value,& what is the problem if a sequence might approach it from positive side?

A Brilliant Member - 7 years, 8 months ago

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As was explained, in your last line of

S ≤ 4 + 1 1 + c 1 + 1 1 + c 2 + 1 1 + c 3 − 3 1 + c 1 c 2 c 3 3 , S \leq 4 + \frac{1}{1+c_1} + \frac{1}{1+c_2} + \frac{1}{1+ c_3} - \frac{3}{1+\sqrt[3]{c_1c_2c_3} },

the RHS is not bounded above by 4 .

In fact ,if c 1 = 8 , c 2 = 1 , c 3 = 1 c_1 = 8, c_2 = 1, c_3 = 1 , we get the value of 4 + 1 9 4 + \frac{1}{9} . In this case, why must we still have S ≤ 4 S \leq 4 , if all we can show is S ≤ 4 + 1 9 S \leq 4 + \frac{1}{9} ?

By taking larger values, c 1 = n 3 , c 2 = c 3 c_1 = n^3, c_2 = c_3 , we get

4 + 1 1 + n 3 + 1 − 3 1 + n 4 + \frac{1}{1+n^3} + 1 - \frac{ 3}{1+n}

which approaches 5 as n n approaches infinity.

Hence, your initial AM-GM bound gave up too much room.

Calvin Lin Staff - 7 years, 8 months ago

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And 1 more thing to all Brilliant staff: Please don't get annoyed if I comment anything trivially wrong on this forum(personally speaking), since most of this I learnt from Brilliant, & essentially this is a learning process. Thanks again Calvin L. & Alexander B.

A Brilliant Member - 7 years, 8 months ago

Correct! Nice explanation of the inner part. I got it Challenge Master & Alexander B. Thanks for clarifications. Well, was this an influence which drove you to give an interesting inequality problem this week (Algebra Level 4)?? I hope yes. :)

A Brilliant Member - 7 years, 8 months ago

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