Can you prove this trigonometric inequality?

[MarkFL]

Show that :

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

 

[Albert1]

Show that :

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

left side=
\(\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}\)
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
Are you sure , right side is correct ?

 

[MarkFL]

left side=
\(\displaystyle \left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}\)
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
Are you sure , right side is correct ?

Yes, it is correct...it appears you are assuming the two sinusoidal factors are in phase with one another, that is for $a=b$. In this case, then your result is equivalent to that which I gave.

 

[Albert1]

if it is correct then ,we must prove
$\dfrac {a^2+b^2}{2}\leq (\dfrac{a+b}{2})^2=\dfrac {a^2+b^2}{4}+{ab}$
for all $a,b \in R$
${\therefore \dfrac {a^2+b^2}{4}\leq ab}$
how about if ab<0,then it does not fit

 

[MarkFL]

if it is correct then ,we must prove
$\dfrac {a^2+b^2}{2}\leq (\dfrac{a+b}{2})^2=\dfrac {a^2+b^2}{4}+{ab}$
for all $a,b \in R$
${\therefore \dfrac {a^2+b^2}{4}\leq ab}$
how about if ab<0,then it does not fit

It appears you are on the right track here, but have made some algebraic errors.

 

[Albert1]

sorry ,I have made some algebraic errors

I will try to use another approach

 

[MarkFL]

sorry ,I have made some algebraic errors

I will try to use another approach

Your errors are quite minor, and in fact leads to a much simpler approach than I have. (Nod)

 

[Albert1]

I think ,I should take a rest ,and have a cup of tea or coffee

 

[Albert1]

for some x,and a,b if left side $\leq 0$
then it holds naturely
now we assume both sides are positive
if a=b then the original inequality holds
if a>b then :$1+b^2\leq left \,\, side \leq 1+a^2$
$1+b^2\leq right \,\, side \leq 1+a^2$
if a<b then :$1+a^2\leq left \,\, side \leq 1+b^2$
$1+a^2\leq right \,\, side \leq 1+b^2$

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[anemone]

My solution:

Spoiler

I first expand the LHS of the inequality and get:

\(\displaystyle ( {\sin x + a\cos x} )( {\sin x + b\cos x})=\sin^2 x+(a+b)\sin x \cos x+ab\cos^2 x\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=(1-\cos^2 x)+(a+b)\sin x \cos x+ab\cos^2 x\)\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{ab+1}{2}+\left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x\)
Next, by applying the Cauchy-Schwarz Inequality to the part \(\displaystyle \left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x\) yields

\(\displaystyle \left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x\le\sqrt{\left(\frac{a+b}{2}\right)^2+\left( \frac{ab-1}{2}\right)^2}\cdot\sqrt{\sin^2 2x+\cos^2 2x}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\le \sqrt{\left(\frac{a^2b^2+a^2+b^2+1}{4}\right)}\)

\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\le \frac{\sqrt{(1+a^2)(1+b^2)}}{2}\)Also, AM-GM inequality tells us that

\(\displaystyle \frac{(1+a^2)+(1+b^2)}{2}\ge\sqrt{(1+a^2)(1+b^2)}\) or

\(\displaystyle \frac{(1+a^2)+(1+b^2)}{4}\ge\frac{\sqrt{(1+a^2)(1+b^2)}}{2}\)

\(\displaystyle \frac{2+a^2+b^2}{4}\ge\frac{\sqrt{(1+a^2)(1+b^2)}}{2}\)Finally, by combining all that we found in the above steps, we can now conclude that

\(\displaystyle ( {\sin x + a\cos x} )( {\sin x + b\cos x})\)

\(\displaystyle =\frac{ab+1}{2}+\left(\frac{a+b}{2}\right)\sin 2x+\left(\frac{ab-1}{2}\right)\cos 2x\)

\(\displaystyle \le \frac{2+a^2+b^2}{4}+\frac{ab+1}{2}\)

\(\displaystyle \le \frac{2+a^2+b^2+2ab+2}{4}\)

\(\displaystyle \le \frac{4+a^2+b^2+2ab}{4}\)

\(\displaystyle \le 1+\frac{a^2+b^2+2ab}{4}\)

\(\displaystyle \le 1+\frac{(a+b)^2}{4}\)

\(\displaystyle \le 1+(\frac{a+b}{2})^2\) (Q.E.D.)

 

[MarkFL]

This is my proof:

Spoiler

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:

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

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

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:

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

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

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

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

\(\displaystyle 4a^2b^2+4a^2+4b^2+4\le a^4+2a^2b^2+4a^2+b^4+4b^2+4\)

\(\displaystyle 2a^2b^2\le a^4+b^4\)

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