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 $u^2$.

This problem was bothering me for a while. The minimum value of $u $ seemed relatively easy to find by using AM-GM followed by Cauchy-Schwarz but both equalities don't hold simultaneously. It seems a bit tempting to assume that the maximum occurs at $x= {\pi/4}$ but I couldn't find a proof to the conjecture. Would someone please help me to find the maximum and minimum with a short relevant proof?

Note: I discovered that the function $u $ is periodic with period ${\pi/2} $.

  • 1
    $\begingroup$ Are there any restrictions on $a$ and $b$? $\endgroup$
    – Vasya
    Aug 17 '19 at 17:31
  • $\begingroup$ They are real constants. The answer depends on them. $\endgroup$ Aug 17 '19 at 17:33
  • $\begingroup$ @Shashwat1337, What about the minimum value? $\endgroup$ Aug 17 '19 at 18:17
  • $\begingroup$ Sorry, in AM-GM and Cauchy Schwatz, both equalities don't hold simultaneously. I've edited the question. $\endgroup$ Aug 19 '19 at 18:04

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

  • 3
    $\begingroup$ Would you please give me another hint? $\endgroup$ Aug 18 '19 at 7:06

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


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

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


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

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


$p + q = a + b$


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

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


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

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

Please check the calculations.

  • $\begingroup$ Thanks a lot. But I was looking for a more elegant solution. Is any such solution possible? $\endgroup$ Aug 20 '19 at 15:44


$$ (a\cos^2x+b\sin^2 x)(b\cos^2x+a\sin^2x) = \frac 18\left((a+b)^2+4ab-(a-b)^2\cos(4x)\right) $$


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