Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$

Question

Let $a, b, c $ be real numbers such that $a<b<c$ and satisfying $$a+b+c=5;$$ $$a^2+b^2+c^2=11;$$ $$a^3+b^3+c^3=27.$$ Prove that $0<a<1<b<2<c<3$.

My understanding :

$(a+b+c)^2 - (a^2+b^2+c^2) = 14$

so, $ab+bc+ca = 7$

$a^2+b^2+c^2-ab-bc-ca = 4$

$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=20$

so, $abc = -\frac{7}{3}$

Thus $a, b, c$ are roots of the equation $P(x), x^3-5x^2+7x-\frac{7}{3}=0$.

We have,

$P(0) = -\frac{7}{3} <0$

$P(1) = 1-5+7-\frac{7}{3} >0$

$P(2) = 8-20+14-\frac{7}{3} <0$

$P(3) = 27-45+21-\frac{7}{3} >0$.

by Intermediate value theorem, $P(x)$ has a root in each of the interval $(0,1), (1, 2), (2,3)$.

Since $a<b<c$, we get $0<a<1<b<2<c<3$.

Michael Rozenberg · Accepted Answer · 2017-07-14T14:08:03.350

3

$$a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)$$ $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+ac+bc)+3abc$$ Get from here $ab+ac+bc$ and $abc$.

I got $ab+ac+bc=7$ and $abc=\frac{7}{3}$, which gives that $a$, $b$ and $c$ are roots of the equation $$x^3-5x^2+7x-\frac{7}{3}=0$$ or $$\frac{1}{(x-1)^3}-\frac{3}{x-1}=-\frac{3}{2}$$ Now, let $\frac{1}{x-1}=2\cos\phi$.

Hence, $\cos3\phi=-\frac{3}{4}$ and the rest is smooth for finding of values $a$, $b$ and $c$, but it's unnecessary.

Let $f(x)=x^3-5x^2+7x-\frac{7}{3}$ and check $f(0)$, $f(1)$, $f(2)$ and $f(3)$.

edited Jul 14 '17 at 14:08

answered Jul 14 '17 at 13:51

Michael Rozenberg

194,933

Thank you, Michael Rozenberg. Is my understanding above correct ? – user403160 Jul 14 '17 at 15:59
1

@carat I think you understood it right. – Michael Rozenberg Jul 14 '17 at 22:49
Thank you very much. Sorry, while my sister was reading your answer on my mobile yesterday. she accidentally down voted your answer. I just knew because I cannot up vote now. :) – user403160 Jul 15 '17 at 00:30
@carat All correct! Thank you! – Michael Rozenberg Jul 15 '17 at 04:13

Jack D'Aurizio · Answer 2 · 2017-07-14T14:16:39.207

This is a classical exercise about symmetric functions. By Newton's identities $$ e_1 = p_1 = 5, $$ $$ 2e_2 = p_1^2 - p_2 = 14, $$ $$ 3e_3 = e_2 p_1 - e_1 p_2 + p_3 = 7 $$ hence $a,b,c$ are roots of the polynomial $$ p(z)=(z-a)(z-b)(z-c) = z^3-e_1 z^2+e_2 z-e_3 = z^3-5z^2+7z-\frac{7}{3} $$ and the claim follows from the fact that $p(0)<0, p(1)>0, p(2)<0, p(3)>0$.
Numerically the roots lie at $ \{0.485131\ldots,1.72235\ldots,2.79252\ldots\} $.

To perform the above manipulations in a single step, one may exploit the $\text{EXP}/\text{LOG}$ map explained here and compute the Taylor series at the origin, up to the cubic term, of $$ \exp\left(-5z+\frac{11}{2}z^2-9z^3\right)=1+5z+7z^2+\frac{7}{3}z^3+\ldots$$ to recover the coefficients of $p(z)$.

Thank you, Jack D'Aurizio. I'll read more to understand it :) — user403160, Jul 14 '17 at 16:12

Solving equations $a+b+c=5,a^2+b^2+c^2=11,a^3+b^3+c^3=27$

2 Answers2