Let $a$,$b$,$c$ be nonnegative reals such that $$a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc=1$$Find the maximum and minimum value of $a+b+c$.
Problem
Source: 2018 China North Mathematical Olympiad Grade 10 Test 1 P2
Tags: inequalities, China, three variable inequality
Nguyenhuyen_AG
26.07.2018 09:38
Devastator wrote: Let $a$,$b$,$c$ be nonnegative reals such that $$a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc=1$$Find the maximum and minimum value of $P=a+b+c$. Note that \[(a+b+c)^2 - \frac{23}{15}\left(a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc\right) = -\frac{(48a-21b-44c)^2+23(9b-4c)^2}{4320}.\]Therefore \[(a+b+c)^2 \leqslant \frac{23}{15}\left(a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc\right) = \frac{23}{15},\]or \[-\sqrt\frac{23}{15} \leqslant a +b +c \leqslant \sqrt\frac{23}{15}.\]Other way, $P=-\sqrt\frac{23}{15}$ when $a = -\frac{2\sqrt{345}}{69},\,b = -\frac{4\sqrt{345}}{345},\, c = -\frac{3\sqrt{345}}{115}.$ And $P=\sqrt\frac{23}{15}$ when $a = \frac{2\sqrt{345}}{69},\,b = \frac{4\sqrt{345}}{345},\, c = \frac{3\sqrt{345}}{115}.$
pankajsinha
25.09.2018 05:37
How is one supposed to get $\frac{23}{15}$ in first place
Plops
14.03.2020 04:09
First note both $(a+b+c)^2$ and $a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc=1$ are homogeneous polynomials of degree $2$. Furthermore, it is enough to find the largest $k$ such that $(a+b+c)^2-k(a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc) \leq 0$.
$$(a+b+c)^2-k(a^2+b^2+c^2+ab+\frac{2}{3}ac+\frac{4}{3}bc)=(1-k)a^2+(1-k)b^2+(1-k)c^2+(2-k)ab+(2-\frac{4}{3}k)bc+(2-\frac{2}{3}k)ab$$Now here is the sucky part, and why I don't really like this problem. To finish it off (as far as I can see), you have to know abut Cholesky decomposition. What we want is to find the largest root $r$ of the coefficients obtained by the Cholesky decomposition of $(1-k)a^2+(1-k)b^2+(1-k)c^2+(2-k)ab+(2-\frac{4}{3}k)bc+(2-\frac{2}{3}k)$. The three coefficients are $$1-k, \frac{\frac{3k^2}{4}-k}{\text{something}}, \frac{(1-k)^2(1-\frac{2}{3}k)+2(1-\frac{k}{2})(1-\frac{k}{3})(1-k)-(1-k)(1-\frac{2}{3}k)^2-(1-\frac{k}{2})^2)(1-k)-(1-k)(1-\frac{1}{3}k)^2}{\text{something}}$$It turns out that the third coefficient has a largest zero of $r=\frac{23}{15}$, and this is the largest zero of all three coefficients. Then we can proceed as @above did.
Definitely wouldn't have gotten this on the actual test. Edit: I seem to have deleted most of my solution. I will correct it in an hour.
Plops
14.03.2020 06:21
Edit: I edited the solution (actually motivation) so that it is full.
sarjinius
07.04.2021 16:14
Note that $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \geq a^2 + b^2 + c^2 + ab + \frac{2}{3}ac + \frac{4}{3}bc = 1 \implies a + b + c \geq \boxed{1}$. Equality holds when $a=1, b=0, c=0$. Let $a_1 = a, b_1 = \frac{b}{2}, c_1 = \frac{c}{3}$, then $1 = a^2 + b^2 + c^2 + ab + \frac{2}{3}ac + \frac{4}{3}bc = a_1^2 + 4b_1^2 + 9c_1^2 + 2a_1b_1 + 2a_1c_1 + 8b_1c_1 = (a_1 + b_1 + c_1)^2 + 3(b_1 + c_1)^2 + 5c_1^2$. Then by Cauchy-Schwarz, $((a_1 + b_1 + c_1)^2 + 3(b_1 + c_1)^2 + 5c_1^2)\left(1 + \frac{1}{3} + \frac{1}{5}\right) \geq (a_1 + 2b_1 + 3c_1)^2 = (a + b + c)^2 \implies (a + b + c)^2 \leq \frac{23}{15} \implies a + b + c \leq \boxed{\frac{\sqrt{345}}{15}}$. Equality holds when $a = \frac{2\sqrt{345}}{69},\,b = \frac{4\sqrt{345}}{345},\, c = \frac{3\sqrt{345}}{115}$.