If $\overrightarrow a$ and $\overrightarrow b$ are unit vectors, then the greatest value of

$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$ is_____.

Let angle between $\overrightarrow a$ and $\overrightarrow b$ be $\theta$. <br><br>$$\sqrt 3 \left| {\overrightarrow a + \overrightarrow b } \right| + \left| {\overrightarrow a - \overrightarrow b } \right|$$ <br><br>= $\sqrt 3 \left( {\sqrt {1 + 1 + 2\cos \theta } } \right)$ + $\left( {\sqrt {1 + 1 - 2\cos \theta } } \right)$ <br><br>= $\sqrt 3 \left( {\sqrt {2 + 2\cos \theta } } \right)$ + $\left( {\sqrt {2 - 2\cos \theta } } \right)$ <br><br>= $\sqrt 6 \left( {\sqrt {1 + \cos \theta } } \right)$ + $\sqrt 2 \left( {\sqrt {1 - \cos \theta } } \right)$ <br><br>= $\sqrt 6 \left( {\sqrt {2{{\cos }^2}{\theta \over 2}} } \right)$ + $\sqrt 2 \left( {\sqrt {2{{\sin }^2}{\theta \over 2}} } \right)$ <br><br>= $2\sqrt 3 \left| {\cos {\theta \over 2}} \right|$ + 2$\left| {\sin {\theta \over 2}} \right|$ <br><br>$\le$ $\sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2}}$ = 4 <br><br><b>Note :</b> |x| = $\sqrt {{x^2}}$ <br><br>|x - 1| = $\sqrt {{{\left( {x - 1} \right)}^2}}$ <br><br>|sin x| = $\sqrt {{{\sin }^2}x}$ <br><br>That is why ${\sqrt {{{\sin }^2}{\theta \over 2}} }$ = $\left| {\sin {\theta \over 2}} \right|$ and ${\sqrt {{{\cos }^2}{\theta \over 2}} }$ = $\left| {\cos {\theta \over 2}} \right|$