$a$,$b$,$c$ are real. What is the highest value of $a+b+c$ if $a^2+4b^2+9c^2-2a-12b+6c+2=0$
Problem
Source: Moldova National Olympiad 2010 9.1
Tags: algebra, Cauchy Inequality
NikoIsLife
17.09.2019 11:19
The equation can also be expressed as
$$(a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2=9$$By Cauchy-Schwarz,
$$\left((a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2\right)\left(1+\frac14+\frac19\right)\ge\left((a-1)+\left(b-\frac32\right)+\left(c+\frac13\right)\right)^2$$$$\frac{49}4\ge\left(a+b+c-\frac{13}6\right)^2$$$$-\frac72\le a+b+c-\frac{13}6\le\frac72$$$$-\frac43\le a+b+c\le\frac{17}3$$This shows that the maximum value of $a+b+c$ is $\boxed{\frac{17}3}$, and equality occurs if and only if $(a,b,c)=\left(\frac{25}7,\frac{15}7,-\frac1{21}\right)$.
Similarly, one can also show that the minimum value of $a+b+c$ is $-\frac43$.
Math-wiz
17.09.2019 11:26
NikoIsLife wrote:
The equation can also be expressed as
$$(a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2=9$$By Cauchy-Schwarz,
$$\left((a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2\right)\left(1+\frac14+\frac19\right)\ge\left((a-1)+\left(b-\frac32\right)+\left(c+\frac13\right)\right)^2$$$$\frac{49}4\ge\left(a+b+c-\frac{13}6\right)^2$$$$-\frac72\le a+b+c-\frac{13}6\le\frac72$$$$-\frac43\le a+b+c\le\frac{17}3$$This shows that the maximum value of $a+b+c$ is $\boxed{\frac{17}3}$, and equality occurs if and only if $(a,b,c)=\left(\frac{25}7,\frac{15}7,-\frac1{21}\right)$.
Similarly, one can also show that the minimum value of $a+b+c$ is $-\frac43$.
Beautiful solution. I liked the first step itself. I didn't realise taking that 4 and 9 outside the bracket
rchokler
18.09.2019 07:11
This equation is an ellipsoid $(a-1)^2+(2b-3)^2+(3c+1)^2=9$. Let $a+b+c=R$ be tangent to the ellipsoid. By affine transformation $(x,y,z)=(a,2b,3c)$. Then we get that plane $6x+3y+2z=6R$ is tangent to the sphere $(x-1)^2+(y-3)^2+(z-1)^2=9$ of radius $3$. The plane has normal vector $\vec{n}=\langle 6,3,2\rangle$ which has length $7$, so the possible tangency points are $(1,3,-1)\pm\frac{3}{7}\vec{n}$ which are $\left(\frac{25}{7},\frac{30}{7},-\frac{1}{7}\right)$ and $\left(-\frac{11}{7},\frac{12}{7},-\frac{13}{7}\right)$. This corresponds to the maximum and minimum $R$. Since the two points give $6x+3y+2z=34$ and $6x+3y+2z=-8$ respectively, we know that $-\frac{4}{3}\leq a+b+c\leq\frac{17}{3}$.
NJOY
18.09.2019 07:38
NikoIsLife wrote:
The equation can also be expressed as
$$(a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2=9$$By Cauchy-Schwarz,
$$\left((a-1)^2+4\left(b-\frac32\right)^2+9\left(c+\frac13\right)^2\right)\left(1+\frac14+\frac19\right)\ge\left((a-1)+\left(b-\frac32\right)+\left(c+\frac13\right)\right)^2$$$$\frac{49}4\ge\left(a+b+c-\frac{13}6\right)^2$$$$-\frac72\le a+b+c-\frac{13}6\le\frac72$$$$-\frac43\le a+b+c\le\frac{17}3$$This shows that the maximum value of $a+b+c$ is $\boxed{\frac{17}3}$, and equality occurs if and only if $(a,b,c)=\left(\frac{25}7,\frac{15}7,-\frac1{21}\right)$.
Similarly, one can also show that the minimum value of $a+b+c$ is $-\frac43$.
Nice solution!!