Find minimum and maximum value of $\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$ 
Blockquote

Find min,max of function $$T=\sqrt{5\cos ^2x+1}+\sqrt{5\sin ^2x+1}$$



*

*Max


By squaring both side and AMGM:
$$T^2=7+2\sqrt{5\cos^2x+1}\cdot \sqrt{5\sin^2x+1}$$ 
$$\le 7+5\left(\cos^2x+\sin^2x\right)+2=14$$
Or $$T\le \sqrt {14}$$


*About minimal value:By $\sqrt x +\sqrt y \ge \sqrt{x+y}$
So $T\ge \sqrt{5+1+1}=\sqrt 7$
I think the minimal value has a little wrong. Pls help me.

Hint. That's equivalent to extremising $$\sqrt{6-5x}+\sqrt {1+5x}$$ on the interval $[0,1].$ Can you now continue?
