Trigonometric Inequality

Show that :

$\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2$

Let:

$\displaystyle A=\tan^{\small{-1}}(a)$

$\displaystyle B=\tan^{\small{-1}}(b)$

Using a linear combination, we may write the inequality as:

$\displaystyle \sqrt{(1+a^2)(1+b^2)}\sin(x+A)\sin(x+B)\le1+\left(\frac{a+b}{2} \right)^2$

Let:

$\displaystyle f(x)=\sin(x+A)\sin(x+B)$

Thus, differentiating with respect to $x$, we obtain::

$\displaystyle f'(x)=\sin(2x+A+B)$

$\displaystyle f''(x)=2\cos(2x+A+B)$

Then $f(x)$ has its maxima for:

$\displaystyle x=\frac{(2k+1)\pi-(A+B)}{2}$ where $\displaystyle k\in\mathbb Z$

We then find:

$\displaystyle f\left(\frac{(2k+1)\pi-(A+B)}{2} \right)=\sin\left(\frac{(2k+1)\pi-(A+B)}{2}+A \right)\sin\left(\frac{(2k+1)\pi-(A+B)}{2}+B \right)=$

$\displaystyle \sin\left(\frac{(2k+1)\pi+A-B}{2} \right)\sin\left(\frac{(2k+1)\pi-A+B}{2} \right)=$

$\displaystyle \frac{\cos(A-B)-\cos((2k+1)\pi)}{2}=\frac{\cos(A-B)+1}{2}=$

$\displaystyle \frac{\cos(A)\cos(B)+\sin(A)\sin(B)+1}{2}=$

$\displaystyle \frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2\sqrt{(1+a^2)(1+b^2)}}$

Now, we need only show:

$\displaystyle \sqrt{(1+a^2)(1+b^2)}f\left(\frac{(2k+1)\pi-(A+B)}{2} \right)\le1+\left(\frac{a+b}{2} \right)^2$

$\displaystyle \frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2}\le1+\left(\frac{a+b}{2} \right)^2$

$\displaystyle 2+2ab+2\sqrt{(1+a^2)(1+b^2)}\le4+a^2+2ab+b^2$

$\displaystyle 2\sqrt{(1+a^2)(1+b^2)}\le2+a^2+b^2$

$\displaystyle 4a^2b^2+4a^2+4b^2+4\le a^4+2a^2b^2+4a^2+b^4+4b^2+4$

$\displaystyle 2a^2b^2\le a^4+b^4$

$\displaystyle 0\le(a^2-b^2)^2$

QED