Find all triples of non-negative real numbers $(a,b,c)$ which satisfy the following set of equations $$a^2+ab=c$$$$b^2+bc=a$$$$c^2+ca=b$$
Problem
Source: RMO 2019 Paper 2 P3
Tags: india, RMO, algebra, Inequality, system of equations
kapilpavase
10.11.2019 16:27
Clearly $(0,0,0)$ is a solution when one of them is $0$... So assume none of them is $0$. We have $\dfrac{c}{a+b}=a$. Summing over and using Nesbitt we have $\sum a\ge3/2$ But we have $$(\sum a)^2-\sum ab=\sum a$$Which implies $$(\sum a)^2-\sum a=\sum ab \le (\sum a)^2/3$$Which yields $\sum a \le 3/2$. Hence equality holds everywhere so $(1/2,1/2,1/2)$ is the solution
BOBTHEGR8
10.11.2019 16:33
Answer is $(a,b,c)=(0,0,0)$ or $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ Proof- If $abc=0$ or if any two are equal we directly get $(a,b,c)=(0,0,0)$or $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$. Now let $a,b,c> 0$ be distinct. Multilply all of the equations to get $(a+b)(b+c)(c+a)=1$ Now subtract them two at a time to get- $a^2-b^2=(c-a)(1+b)$ $b^2-c^2=(a-b)(1+c)$ $c^2-a^2=(b-c)(1+a)$ Multiply all these to get $1=(a+b)(b+c)(c+a)=(1+a)(1+b)(1+c)$ But $a,b,c>0$ . Contradiction. Hence proved.
char2539
10.11.2019 17:18
Here is a solution which directly falls to extremal principle
Trivially $(0,0,0)$ is a solution.Henceforth assume none of $a,b,c$ are equal to $0$
Assume WLOG $c \geq b \geq a$.From the first equation we get $2b^2 \geq c \geq a$ and from the second equation we get $a \geq 2b^2$.So $a=b=c$ and hence the other solution is $(1/2,1/2,1/2)$and we are done.
kapilpavase
10.11.2019 17:36
What would you do in the case $c\ge a\ge b$
char2539
10.11.2019 17:41
@above ig its the same thing and its just we get $2a^2 \geq c \geq b \geq 2a^2$ Actually i should have stated like "order the a,b,c"in order to avoid the confusion.
Al3jandro0000
10.11.2019 18:20
BOBTHEGR8 wrote: Answer is $(a,b,c)=(0,0,0)$ or $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ Proof- If $abc=0$ or if any two are equal we directly get $(a,b,c)=(0,0,0)$or $(\frac{1}{2},\frac{1}{2},\frac{1}{2})$. Now let $a,b,c> 0$ be distinct. Multilply all of the equations to get $(a+b)(b+c)(c+a)=1$ Now subtract them two at a time to get- $a^2-b^2=(c-a)(1+b)$ $b^2-c^2=(a-b)(1+c)$ $c^2-a^2=(b-c)(1+a)$ Multiply all these to get $1=(a+b)(b+c)(c+a)=(1+a)(1+b)(1+c)$ But $a,b,c>0$ . Contradiction. Hence proved. What implies the contradiction?
BOBTHEGR8
10.11.2019 18:27
Al3jandro0000 wrote: What implies the contradiction? $(1+a)(1+b)(1+c)=1$ and $ a,b,c>0$
math_and_me
10.11.2019 18:56
#4 the system of equalities is cyclic and not symmetric in $a,b,c$. So you have to take two cases while assuming any ordering.
rchokler
10.11.2019 20:29
And if solved in complex numbers instead? $(0,0,0)$, $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$, and cyclic permutations of $(A,B,C)$, $(\bar{A},\bar{B},\bar{C})$. $A=-\frac{1}{6}+\frac{i\sqrt{3}}{6}+\frac{1+i\sqrt{3}}{12}\sqrt[3]{28+12 i \sqrt{3}}+\frac{1}{6} \sqrt[3]{44+84 i \sqrt{3}}$ $B=-\frac{1}{6}+\frac{i\sqrt{3}}{6}-\frac{1}{6}\sqrt[3]{28+12 i \sqrt{3}}-\frac{1+i\sqrt{3}}{12} \sqrt[3]{44+84 i \sqrt{3}}$ $C=-\frac{1}{6}+\frac{i\sqrt{3}}{6}+\frac{1-i\sqrt{3}}{12}\sqrt[3]{28+12 i \sqrt{3}}-\frac{1-i\sqrt{3}}{12} \sqrt[3]{44+84 i \sqrt{3}}$
whizkid32
11.11.2019 12:07
how to solve this when a,b,c range over all real numbers
BOBTHEGR8
20.11.2019 06:31
@above I think @2above has given solution of a more general case.
Keith50
08.08.2020 18:24
The answers are $(a,b,c)=(0,0,0), \left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$. It is easy to see that these are solutions. If $a,b,c\ne 0$, we have \[(a+b)(b+c)(c+a)=\frac{c}{a}\cdot \frac{a}{b}\cdot \frac{b}{c}=1.\]Now let $u=a+b,v=b+c,w=c+a$, we then have \[a=\frac{u-v+w}{2}, b=\frac{u+v-w}{2}, c=\frac{-u+v+w}{2}\], so rewriting gives us \[u^2+uw-uv=-u+v+w\]\[v^2+uv-vw=u-v+w\]\[w^2+vw-uw=u+v-w\]and adding up gives \[u^2+v^2+w^2=u+v+w\]but recalling that $uvw=1$, by Muirhead, \[u^2+v^2+w^2\geq \sqrt[3]{uvw}(u+v+w)\]which equality occurs when $u=v=w$ and this is our case. So, this implies $a+b=b+c=c+a\implies a=b=c=\frac{1}{2}$.
SatisfiedMagma
30.01.2021 11:57
kapilpavase wrote: Clearly $(0,0,0)$ is a solution when one of them is $0$... So assume none of them is $0$. We have $\dfrac{c}{a+b}=a$. Summing over and using Nesbitt we have $\sum a\ge3/2$ But we have $$(\sum a)^2-\sum ab=\sum a$$Which implies $$(\sum a)^2-\sum a=\sum ab \le (\sum a)^2/3$$Which yields $\sum a \le 3/2$. Hence equality holds everywhere so $(1/2,1/2,1/2)$ is the solution Wow! You remembered Nesbitt? Which inequalities you remember as lemma for olympiads? My solution was rather a bash...
508669
30.01.2021 14:57
Math-wiz wrote: Find all triples of non-negative real numbers $(a,b,c)$ which satisfy the following set of equations $$a^2+ab=c$$$$b^2+bc=a$$$$c^2+ca=b$$ It can be seen that if $a=0$ then $c=0$ using first equation and using second equation we have that $b=0$, so let us say that $a \neq b \neq c = 0$. Let $a \geq b, c$. Therefore, $a+b > c$ and $a(a+b)=c$ and so $a < 0$, implying that $(a, b, c) < 1$. However, we notice that if $a > 0.5$, then $c = a^2 + ab > 0.25 + 0.5b$ and $b = c^2 + ac > c^2 + 0.5c = c(c+0.5) > (0.25+0.5b)(0.75+0.5b) = 0.25b^2 + 0.25 \cdot 0.75 + 0.5b$ implying that $b > 0.375 + 0.5b^2$ or $b > 0.5$ and so $c > 0.5$, but multiplying all equation give that $(a+b)(b+c)(c+a) = 1$ which is a straight contradiction since $a+b, b+c, c+a > 1$, therefore $a \leq 0.5$, but then $(a+b)(b+c)(c+a) = 1 \implies a = b = c = 0.5$ since $a+b, b+c, c+a \leq 1$ with equality only at $a = b = c = 0.5$, therefore only solutions are $\boxed{(a, b, c) = (0, 0, 0), (0.5, 0.5, 0.5)}$
homotopygroup
05.02.2021 11:31
Clearly $(a,b,c)=(0,0,0)$ admits the system trivially, so we assume $a,b,c \neq 0$.. Let's define $\sigma_j$ as elementary symmetric polynomials
Then adding the equations of the system we get $$\sigma_1^{2} -\sigma_2=\sigma_1 \Leftrightarrow \sigma_2 =\sigma_{1}(\sigma_1 -1)$$then multiplying the equations we get $$\prod_{cyc}(\sigma_1 -a)=1 \Leftrightarrow \sigma_3 =\sigma_1 \cdot \sigma_2-1$$Then using Maclaurin's inequalities and the previous results we get $\frac{\sigma_1}{3} \ge \sqrt{\frac{\sigma_2}{3}} \Leftrightarrow \sigma_1 \le \frac{3}{2}$
Then using AM-GM inequality we get $$1=\prod_{cyc}(a+b) \le \left(\frac{2\sigma_{1}}{3} \right)^3 \Leftrightarrow \sigma_1 \ge \frac{3}{2}$$Thus we can conclude that $\sigma_1 =\frac{3}{2}$ , using previous results we get $\sigma_2 =\frac{3}{4}$ and $\sigma_3 =\frac{1}{8}$ , as $(a,b,c)=\left(\frac{1}{2},\frac{1}{2}, \frac{1}{2} \right)$ is the only triple in $\mathbb{R}^{3}$ satisfying the desired conditions we can conclude it to be the only non-zero solution
lifeismathematics
24.07.2022 16:42
notice $\Sigma a^2+\Sigma ab=\Sigma a$, so $(\Sigma a)^2-\Sigma a=\Sigma ab$
also as $ab=c-a^2\\
bc=a-b^2\\
ca=b-c^2$
so we have $\Sigma ab=\Sigma a-\Sigma a^2$, which gives $(\Sigma a)^2-\Sigma a=\Sigma a-\Sigma a^2$.
and from a very fundamental inequality $(a+b+c)^2 \le 3(a^2+b^2+c^2)$. We have $(\Sigma a)^2 \le 6\Sigma a-3(\Sigma a)^2$. Which gives that $0\le \Sigma a \le \frac{3}{2}$.
also notice $(a+b)(b+c)(c+a)=1$.(this can be easily formulated by substracting simultaneously the equation, and multiplying each), now from AM-GM on $(a+b),(b+c),(c+a)$ we get $\frac{2(a+b+c)}{3}\ge 1$ , so we get $a+b+c \ge \frac{3}{2}$. So we see that $a+b+c=\frac{3}{2}$ or equality holds hence $a=b=c=\frac{1}{2}$. One can trivially notice that 0,0,0 is a trivial solution hence $\boxed{(0,0,0) ,(\frac{1}{2},\frac{1}{2},\frac{1}{2})}$ as our class of answers
Jishnu4414l
09.10.2024 14:32
Let no two of $(a,b,c)$ be equal. Clearly, we have $(a+b)(b+c)(c+a)=(a+1)(b+1)(c+1)=1$. Clearly, $(0,0,0)$ is the only solution in non-negative reals. However, this is a contradiction because we assumed $a\neq b\neq c$. Now, if any two of $a,b,c$ are equal, the third is equal as well. Let $a=b=c=k$. Multiplying the three equations, we get that $8k^6=k^3$. The only solutions are $k=0$ or $k=2$. These values of $k$ correspond to $(a,b,c)=(\frac{1}{2},\frac{1}{2},\frac{1}{2})$ or $(0,0,0)$. We are done!