I stumbled across this problem which goes as follows:
given |z-3| + |z| + |z+3| = 12 ,we need to find maximum and minimum value of [|z|] where [a] is greatest integer less than or equal to a.
my approach:
well for expressions like |z-b| + |z-c| = k it is easy to observe geometrically the locus as ellipse,hyperbola,pair of rays,line segment etc. and then |z| is distance of z from origin so optimizing is easy. however here I don't know the geometric interpretation of |z-3| + |z| + |z+3| = 12 so the only thing I am left with is put z = x+ iy but it still requires atleast 3 rounds of squaring and I could not optimize it that way.
hence I am stuck.Is there any geometrical essence to this expression? what is the best way to optimize this? kindly help me out.
To get the answer without proof, this is kind of like an ellipse (in fact, it's a 3-ellipse) with an axis on the $x$-axis. So it's intuitively obvious we'll have extremes when $x=0$ or $y=0$. Below is a proof of the answer.
Particular points
Note that if $x=0$ and $y>0$ we have $12=|z-3|+|z|+|z+3|=|y|+2\sqrt{9+y^2}=y+2\sqrt{9+y^2}$. Thus, $(12-y)^2=4(9+y^2)$ so $144-24y+y^2=36+4y^2$, so that $y^2+8y-36=0$. Combined with $y>0$ we have $y=2(\sqrt{13}-2)<2(\sqrt{16}-2)=4$ so that $|z|<4$ at this point.
And if $y=0$ and $x>0$ we have $12=|x-3|+x+x+3$. If there is a solution with $x\ge3$, then $12=3x$ and we must have $x=4$, which works. So we have $|z|=4$ at this point.
Upper bound and max
Note that the triangle inequality tells us that $12=|z-3|+|z|+|z+3|\ge|z|+|(z-3)+(z+3)|=|z|+|2z|=3|z|$, so that $|z|\le 4$ no matter what. Therefore, $z=4$ attains the maximum $[|z|]=4$.
Lower bound and min
I suspect there is a more elegant way to show the minimum, but I couldn't find one in a reasonable amount of time.
We will show that $12=|z-3|+|z|+|z+3|$ does not intersect the disk $|z|\le3$, by finding where $|z-3|+|z|+|z+3|$ is maximized on the disk and that the maximum value is less than $12$.
Note that the The $y$-partial of $|z-3|+|z|+|z+3|$ is $\dfrac{y}{|z-3|}+\dfrac{y}{|z|}+\dfrac{y}{|z+3|}$ $=y*\dfrac{|z+3||z|+|z-3||z+3|+|z-3||z|}{|z-3||z||z+3|}$. So a potential critical point where this is $0$ or undefined has to have $y=0$ or $z$ at $0$ or $\pm3$ (all of which have $y=0$ anyway), since the three numerator terms are all nonnegative and can't be $0$ simultaneously.
Since we optimize by looking at critical points or the boundary, $|z-3|+|z|+|z+3|$ is maximized over the disk when $y=0$ or when $|z|=3$, In the former case, $|z-3|+|z|+|z+3|=|x|+3-x+x+3=|x|+6\le|z|+6\le3+6=9$. In the latter case, we can use $(x,y)=3(\cos t,\sin t)$ to reduce to a single variable optimization or Lagrange Multipliers. Either way, we find that we need to look at $(\pm3,0)$ and $(0,\pm3)$. We already looked at the points on the $x$-axis, and at $(0,\pm3)$ we have $|z-3|+|z|+|z+3|=2*3\sqrt{2}+3<6*\frac32+3=12$.
Therefore, $|z|\le3$ is impossible, so the lowest possible value of $[|z|]$ is $3$, occurring when $|z|<4$. We found a point with $|z|<4$ already, so the minimum value is indeed $3$.