<s>[INST] 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.
 [/INST] The quantity $T(x)\geq0$ is a ${\pi\over2}$-periodic function of $x$. It is extremal, i.e., maximal or minimal, iff $$T^2=7+2\sqrt{(5\cos^2 x+1)(5\sin^2 x+1)}$$ is extremal, and this is the case iff the radicand on the RHS
is extremal. Letting $\sin^2 x=:t$ this means that
$$\phi(t):=\bigl(5(1-t)+1\bigr)(5t+1)=6+25(t-t^2)\qquad(0\leq t\leq1)$$
should be extremal. This $\phi(t)$ takes its minimum $6$ at $t=0$ and $t=1$, and takes its maximum $6+{25\over4}={49\over4}$ at $t={1\over2}$.
From $T^2=7+2\sqrt{6}=\bigl(1+\sqrt{6}\bigr)^2$ it follows that the minimum of $T$ is $1+\sqrt{6}$. Similarly, from $$T^2=7+2\cdot\sqrt{49\over4}=14$$ it follows that the maximum of $T$ is $\sqrt{14}$.
</s>