If $u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$ , find the maximum and minimum value of $u^2$.

Put $$1-\cos2x =2\sin^2x$$ $$1+\cos2x =2\cos^2x$$

$u=\sqrt {a\cos^{2}x+b\sin^{2}x} + \sqrt {b\cos^{2}x+a\sin^{2}x}$

Let

$p = a \cos^2 x + b \sin^2 x$

$q = b \cos^2 x + a \sin^2 x$

and

$u = \sqrt{p} + \sqrt{q}$

Then $u^2 = p + q + 2 \sqrt{pq}$

Now

$p + q = a + b$

and

$\displaystyle pq = \frac{(a+b)^2}{4} - \frac{(a-b)^2}{8} - \frac{(a-b)^2}{8} \cos 4x$

If $\cos 4x = -1$ then $u^2$ is maximum and is equal to

$2(a+b)$

If $\cos 4x = 1$ then $u^2$ is minimum and is equal to

$a+b + 2 \sqrt{ab}$

Please check the calculations.