Prove that minimum of $\lambda \sin \theta + (1 - \lambda) \cos \theta \le -\frac1{\sqrt 2}$
Solution 1:
As per @Aryabhata's suggestion, adding the answer from the earlier comment to close this question.
The last part of the question resolves to,
Show that minimum of $-\sqrt {2\lambda^2 - 2\lambda + 1} \le -\dfrac{1}{\sqrt 2}$
$2\lambda^2 - 2\lambda + 1 $ is a parabola whose minimum is at its vertex, $-\dfrac{b}{2a} = \dfrac{-(-2)}{2(2)} = \dfrac{1}{2}$
And the maximum of $-\sqrt {2\lambda^2 - 2\lambda + 1}$ is $-\dfrac{1}{\sqrt 2}$
Hence,
$$ -\sqrt {2\lambda^2 - 2\lambda + 1} \le -\dfrac{1}{\sqrt 2} $$