The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$

My try:I have just normally squared the expression  and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$ 
$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$ 
I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?
 A: Write 
$$\cos^2x=\frac{1+\cos 2x}{2}$$
and
$$\sin^2x=\frac{1-\cos 2x}{2}$$
Then, we have
$$u=\sqrt{A+B\cos 2x}+\sqrt{A-B\cos 2x}\tag 1$$
where 
$$A=\frac{a^2+b^2}{2}$$
$$B=\frac{a^2-b^2}{2}$$
Taking the derivative of u in $(1)$ and setting the derivative equal to zero reveals 
$$\frac{-B\sin 2x}{\sqrt{A+B\cos 2x}}+\frac{B\sin 2x}{\sqrt{A-B\cos 2x}}=0$$
whereupon solving reveals that either $\sin 2x=0$ or $\cos 2x=0$.  When $\cos 2x=0$, 
$$\bbox[5px,border:2px solid #C0A000]{u=\sqrt{2(a^2+b^2)} \,\,\text{is the maximum}}$$
and when $\sin 2x =0$, 
$$\bbox[5px,border:2px solid #C0A000]{u=|a|+|b|\,\,\,\text{is the minimum}}$$ 
A: Expanding $u^2$ more:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^4+b^4)+a^2b^2(\sin^4x+\cos^4x)}$$
Using trigonometric identity $\sin^2x+\cos^2x=1$ we can derive that:$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$$
Rewrite $u^2$ again:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^2-b^2)^2+a^2b^2}$$
The minimum value of $\sin^2x\cos^2x$ is $0$ and its maximum value is (using AM-GM) $$\frac{\sin^2x+\cos^2x}{2}=\frac{1}{2}\ge\sin x\cos x$$
$$\frac{1}{4}\ge\sin^2x\cos^2x$$
Also you can find it this way using trigonometric identities $$\sin^2x\cos^2x = \frac{\sin^2(2x)}{4}\Rightarrow \max(\sin^2x\cos^2x)=\max \left(\frac{\sin^2(2x)}{4}\right)=\frac{1}{4}$$ 
So $$u^2_{min}=(\left |a\right |+\left |b\right |)^2$$
$$u^2_{max}=2(a^2+b^2)$$
A: Assume WLOG $a > b > 0$, $A = \sqrt{a^2\cos^2x+b^2\sin^2x}, B = \sqrt{a^2\sin^2x+b^2\cos^2x}\Rightarrow A^2+B^2 = a^2+b^2\Rightarrow u^2 = (1\cdot A+1\cdot B)^2\leq (1^2+1^2)(A^2+B^2)=2(a^2+b^2)\Rightarrow u^2_{max} = 2(a^2+b^2)$. To find $u^2_{min}$, you need to find the min of $(a^2\cos^2x+b^2\sin^2x)(a^2\sin^2x+b^2\cos^2x)=f(\cos^2 x)=(a^2-(a^2-b^2)t)(b^2+(a^2-b^2)t), t = \cos^2x, 0 \leq t \leq 1=f(p) = (a^2-p)(b^2+p), p = (a^2-b^2)t, 0 \leq p \leq a^2-b^2\to f(p) = a^2b^2 + (a^2-b^2)p - p^2\Rightarrow f'(p) = a^2-b^2 - 2p=0 \iff p = \dfrac{a^2-b^2}{2}\Rightarrow f\left(\dfrac{a^2-b^2}{2}\right)=\dfrac{(a^2+b^2)^2}{4}$. At end points $p = 0, a^2-b^2, f(0) = a^2b^2, f(a^2-b^2) = a^2b^2$. Thus $u^2_{min} = a^2+b^2 + 2\sqrt{a^2b^2}=(a+b)^2$, since $a^2b^2 \leq \dfrac{(a^2+b^2)^2}{4}$.
A: For $\min$
Using $\triangle$ Inequality::
Let $z_{1} = a\cos x+i b\sin x$ and $z_{2} = b\cos x+i a\sin x$
So $$|z_{1}|+|z_{2}|\geq |z_{1}+z_{2}|$$
So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\geq \sqrt{(a+b)\cos^2 x+(a+b)^2\sin^2 x}=|a+b|$$
For $\max$ Same as Deepsea
$$\left[\left(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}\right)^2+\left(\sqrt{a^2\sin^2 x+b^2 \cos^2 x}\right)^2\right]\cdot \left[1^2+1^2\right]\geq \bigg(\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\bigg)^2$$
So $$\sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$
So $$|a+b|\leq \sqrt{a^2\cos^2 x+b^2 \sin^2 x}+\sqrt{a^2 \sin^2 x+b^2 \cos^2 x}\leq \sqrt{2(a^2+b^2)}$$
