MHB Can you prove this trigonometric inequality?

AI Thread Summary
The discussion centers around proving the trigonometric inequality involving sine and cosine functions. Participants analyze the left side of the inequality, suggesting it can be bounded by the product of square roots of expressions involving constants a and b. There is a debate about the correctness of the right side and whether assumptions about a and b being equal affect the proof. Some participants acknowledge algebraic errors in their reasoning and propose alternative approaches to the proof. The conversation highlights the complexity of the inequality and the need for careful consideration of cases where a and b may be negative.
MarkFL
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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]
 
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MarkFL said:
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]
left side=
[math]\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}[/math]
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
Are you sure , right side is correct ?
 
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Albert said:
left side=
[math]\left( {\sin x + a\cos x} \right)\left( {\sin x + b\cos x} \right)\leq \sqrt{1+a^2}\times \sqrt{1+b^2}[/math]
$\leq\dfrac{1+a^2+1+b^2}{2}=1+\dfrac {a^2+b^2}{2}$
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.
 
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
 
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Albert said:
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.
 
sorry ,I have made some algebraic errors:o

I will try to use another approach
 
Albert said:
sorry ,I have made some algebraic errors:o

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)
 
I think ,I should take a rest ,and have a cup of tea or coffee
 
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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  • #10
My solution:

I first expand the LHS of the inequality and get:

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

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

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

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

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

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

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

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

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

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

$$\le \frac{2+a^2+b^2}{4}+\frac{ab+1}{2}$$

$$\le \frac{2+a^2+b^2+2ab+2}{4}$$

$$\le \frac{4+a^2+b^2+2ab}{4}$$

$$\le 1+\frac{a^2+b^2+2ab}{4}$$

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

$$\le 1+(\frac{a+b}{2})^2$$ (Q.E.D.)
 
  • #11
This is my proof:

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:

[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]
 

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