Show that :

[math]\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right) \leq 1 + \left( \frac{a + b}{2} \right)^2[/math]

Let:

[math]A=\tan^{\small{-1}}(a)[/math]

[math]B=\tan^{\small{-1}}(b)[/math]

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

[math]\sqrt{(1+a^2)(1+b^2)}\sin(x+A)\sin(x+B)\le1+\left(\frac{a+b}{2} \right)^2[/math]

Let:

[math]f(x)=\sin(x+A)\sin(x+B)[/math]

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

[math]f'(x)=\sin(2x+A+B)[/math]

[math]f''(x)=2\cos(2x+A+B)[/math]

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

[math]x=\frac{(2k+1)\pi-(A+B)}{2}[/math] where [math]k\in\mathbb Z[/math]

We then find:

[math]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)=[/math]

[math]\sin\left(\frac{(2k+1)\pi+A-B}{2} \right)\sin\left(\frac{(2k+1)\pi-A+B}{2} \right)=[/math]

[math]\frac{\cos(A-B)-\cos((2k+1)\pi)}{2}=\frac{\cos(A-B)+1}{2}=[/math]

[math]\frac{\cos(A)\cos(B)+\sin(A)\sin(B)+1}{2}=[/math]

[math]\frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2\sqrt{(1+a^2)(1+b^2)}}[/math]

Now, we need only show:

[math]\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[/math]

[math]\frac{1+ab+\sqrt{(1+a^2)(1+b^2)}}{2}\le1+\left(\frac{a+b}{2} \right)^2[/math]

[math]2+2ab+2\sqrt{(1+a^2)(1+b^2)}\le4+a^2+2ab+b^2[/math]

[math]2\sqrt{(1+a^2)(1+b^2)}\le2+a^2+b^2[/math]

[math]4a^2b^2+4a^2+4b^2+4\le a^4+2a^2b^2+4a^2+b^4+4b^2+4[/math]

[math]2a^2b^2\le a^4+b^4[/math]

[math]0\le(a^2-b^2)^2[/math]

QED

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