Find minimum of $\frac{x^2}{2}+x-\lambda x$ when $x \geq 1, \ \lambda \geq 0$

Question

The question is quite straight-forward, how does one find the minimum function (depending on $\lambda$) of

$\frac{x^2}{2}+x-\lambda x$ when $x \geq 1, \ \lambda \geq 0$?

I know the answer is $\frac{3}{2} - \lambda$ when $0 \leq \lambda \leq 2$ and $-\frac{(1-\lambda)^2}{2}$ when $\lambda \geq 2$, but how does one compute that?

Answer 1

Note that $$ x^2/2 +(1-\lambda)x = \frac{1}{2}\left((x+(1-\lambda))^2 - (1-\lambda)^2\right). $$ From here, I believe you can derive the answer you want.

Answer 2

I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,\lambda) = \frac{x^{2}}{2}+x - \lambda x$ with constraints $x\geq1$ and $\lambda\geq0$. Now, arrange the function as follows : \begin{align*} f(x,\lambda) &= \frac{1}{2}(x^{2} +2(1-\lambda)x)\\ &= \frac{1}{2}\bigg((x-(1-\lambda))^{2}-(1-\lambda)^{2}\bigg) \end{align*} Hence, $f$ attains minimum whenever $x=1-\lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.