Prove two inequalities. How can we prove these two inequalities, I proved the first question which I guess will be used to prove the following inequalities, but I don't know how to start.  Let $a,b,c$ be there real numbers. Prove that if :
$\sin a+ \sin b + \sin c\ge 2
\implies \cos a+ \cos b + \cos c\le \sqrt 5$
and
$\sin a+ \sin b + \sin c\ge \frac 32\implies \sin (a-\pi/6)+ \sin(b-\pi/6) + \sin (c-\pi/6)\ge 0.$
The first question was to prove that if
$x,y,z$ are real numbers, then $(x+y+z)^2\le 3(x^2+y^2+z^2)$. I proved this part, it is a direct result using $x^2+y^2\ge 2xy$ and by expanding $(x+y+z)^2$. Thanks for your help.
 A: Part 1:
Let $0\le k\le 3$ be any given real number then $\begin{align} & \sin a+ \sin b + \sin c\ge k\\
\implies &  (\sin a+ \sin b + \sin c)^2 \ge k^2\\
\implies & 3(\sin^2 a + \sin^2 b +\sin^2 c) \ge (\sin a+ \sin b + \sin c)^2 \ge k^2 \tag{1}\\
\implies & 3(1-\cos^2 a + 1-\cos^2 b +1- \cos^2 c) \ge k^2\\
\implies & 9-3(\cos^2 a +\cos^2 b + \cos^2 c) \ge k^2\\
\implies & 3(\cos^2 a +\cos^2 b + \cos^2 c) \le 9-k^2\\
\implies & (\cos a + \cos b +\cos c)^2\le 3(\cos^2 a +\cos^2 b + \cos^2 c) \le 9-k^2 \tag{2}\\
\implies &-\sqrt{9-k^2} \le \cos a + \cos b +\cos c \le \sqrt{9-k^2} \tag{3}\end{align}$
where in $(1)$ and $(2)$ we used $(x+y+z)^2\le 3(x^2+y^2+z^2)$
For your first question put $k=2$ and you'll get $\sin a+ \sin b + \sin c\ge 2 \implies \cos a + \cos b +\cos c \le \sqrt{9-2^2}=\sqrt{5}$

Part 2:
Given that $\sin a+ \sin b + \sin c\ge \frac{3}{2} \tag{4}$
We get, $\cos a + \cos b +\cos c \le \sqrt{9-\left(\frac{3}{2}\right)^2}={3\sqrt{3}\over {2}}$ using $(3)$
This implies $-\frac{1}{2} (\cos a + \cos b +\cos c) \ge -{3\sqrt{3}\over {4}} \tag{5}$
Now use $(4)$ and $(5)$ to prove  $\sin (a-\pi/6)+ \sin(b-\pi/6) + \sin (c-\pi/6)\ge 0$
