Infimum taken over $\lambda$ in $\mathbb{C}$ I want to calculate the infimum of 
$$ |\lambda-2|^2+|2\lambda-1|^2+|\lambda|^2 $$
over $\lambda\in\mathbb{C}.$
I choose $\lambda=2,1/2,0$ so that one term in the above expression becomes zeros and minimum comes out $5/2$ at $\lambda =1/2.$ Is this infimum also, please help.
 A: Setting one of the summand to minimum, may not minimize the sum
Let $\lambda = x+iy$
$$F=(x-2)^2+y^2+(2x-1)^2+(2y)^2+(x)^2+(y)^2$$
$$=6x^2-8x+5+6y^2$$
$$\ge6x^2-8x+5$$
Again $6x^2-8x+5=6\left(x-\dfrac23\right)^2+5-\left(\dfrac23\right)^2\ge5-6\cdot\left(\dfrac23\right)^2$
So, the minimum value$\left[5-6\cdot\left(\dfrac23\right)^2\right]$ of $F$ occurs if $y=0,x-\dfrac23=0$
A: If we take $\lambda=x+i y$, we get (as commented above)
\begin{equation}
f(x,y)=6x^2+6y^2-8x+6.
\end{equation}
BUT, to be accurate, then we need to calculated the gradient, so we differentiate by $x$ and by $y$
\begin{alignat}{2}
f_x=12x-8 ~,\\
f_y=12y ~.
\end{alignat}
Hence we get a stationary point $(2/3,0)$. However, to check, if it is an infimum, we need to calculated the second derivative of $f(x,y)$ at that point
\begin{alignat}{2}
\mathrm{d}^2f&={\partial^2 f\over\partial x^2}\mathrm{d}x^2+2{\partial^2 f\over\partial x\partial y}\mathrm{d}x\mathrm{d}y+{\partial^2 f\over\partial y^2}\mathrm{d}y^2=\\
&=12\mathrm{d}x^2+12\mathrm{d}y^2>0~.
\end{alignat}
As $\mathrm{d}^2f>0$, the point $(2/3,0)$ is a minimum, hence $\lambda=2/3$ is an infimum.
A: Note that the given equation is the same as
$$|\lambda-2|^2+4|\lambda-\tfrac{1}{2}|^2+|\lambda^2|.$$
So we are looking for $\lambda$ to minimize the weighted sum of the distances to the points $2,\tfrac{1}{2},0\in\Bbb{C}$, with respective weights $1$, $4$ and $1$. This is the same as the weighted average of the points, which is
$$\frac{1}{1+4+1}(1\cdot2+4\cdot\tfrac{1}{2}+1\cdot0)=\frac{1}{6}(2+2)=\frac{4}{6}=\frac{2}{3}.$$
