Let $$ S := \left\{ (x , y, z) \in \mathbb{R}^3 : x^2 + 2y^2 + 3z^2 \leq 6, x \geq 0 \right\} $$ Prove that $S$ is convex.
I have chosen two points: $\mathbf{(x_1, y_1, z_1)}$ and $\mathbf{(x_2, y_2, z_2)}$ $\in$ S.
And utilized the "line segment technique": $\lambda$x + (1 - λ)y, where $\lambda$ $\in$ [0,1]
In this case ($\mathbb{R}^3$), it gives:
$\lambda$$x_1$+(1-$\lambda$)$x_2$, $\lambda$$y_1$+(1-$\lambda$)$y_2$, $\lambda$$z_1$+(1-$\lambda$)$z_2$
Plugging into S:
($\lambda$$x_1$+(1-$\lambda$)$x_2$)$^2$ + 2($\lambda$$y_1$+(1-$\lambda$)$y_2$)$^2$ + 3($\lambda$$z_1$+(1-$\lambda$)$z_2$)$^2$
Expanding:
($\lambda$$^2$$x_1^2$+2(1-$\lambda$)$\lambda$$x_1$$x_2$+(1-$\lambda$)$^2$$x_2^2$) + 2($\lambda$$^2$$y_1^2$+2(1-$\lambda$)$\lambda$$y_1$$y_2$+(1-$\lambda$)$^2$$y_2^2$) + 3($\lambda$$^2$$z_1^2$+2(1-$\lambda$)$\lambda$$z_1$$z_2$+(1-$\lambda$)$^2$$z_2^2$)
Common factors give:
$\lambda$$^2$($x_1^2$+2$y_1^2$+3$z_1^2$) + 2$\lambda$(1-$\lambda$)($x_1$$x_2$+2$y_1$$y_2$ + 3$z_1$$z_2$) + (1-$\lambda$)$^2$($x_2^2$+2$y_2^2$+3$z_2^2$)
From S, it follows that ($x_1^2$+2$y_1^2$+3$z_1^2$) and ($x_2^2$+2$y_2^2$+3$z_2^2$) $\leq$ 6, which gives
$\lambda$$^2$(6) + 2$\lambda$(1-$\lambda$)($x_1$$x_2$+2$y_1$$y_2$ + 3$z_1$$z_2$) + (1-$\lambda$)$^2$(6)
If we also assume that ($x_1$$x_2$+2$y_1$$y_2$ + 3$z_1$$z_2$) $\leq$ 6, we get
$\lambda$$^2$(6) + 2$\lambda$(1-$\lambda$)(6) + (1-$\lambda$)$^2$(6)
Solving this gives:
6, which is $\leq$ 6 and thus the set needs to be convex. But here my problem arises:
How can we know that ($x_1$$x_2$+2$y_1$$y_2$ + 3$z_1$$z_2$) is indeed $\leq$ 6?
Any help appreciated!
Apply the Cauchy-Schwarz inequality in $\mathbb{R}^3$ : $(x_1 x_2+2y_1y_2 + 3z_1z_2)= \\ \begin{pmatrix}x_1\\ \sqrt{2}y_1 \\ \sqrt{3}z_1 \end{pmatrix} \begin{pmatrix}x_2\\ \sqrt{2}y_2 \\ \sqrt{3}z_2 \end{pmatrix} \leq \sqrt{(x_1)^2 +(\sqrt{2}y_1)^2+(\sqrt{3}z_1)^2} \sqrt{(x_2)^2 +(\sqrt{2}y_2)^2+(\sqrt{3}z_2)^2} \\ = \sqrt{x_1^2 +2y_1^2+3z_1^2} \sqrt{x_2^2 +2y_2^2+3z_2^2} \leq \sqrt{6} \sqrt{6} = 6 $