Prove that (1+a/sinx)(1+b/cosx) is greater than or equal to (1+2sqrt(ab))^2

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In summary, for all real numbers $a, b, x$ with $a, b \geq 0$ and $0 < x < \frac{\pi}{2}$, it has been proven that $(1+\frac{a}{\sin x})(1+\frac{b}{\cos x}) \geq (1+ \sqrt {2ab})^2$ by using the Cauchy-Schwarz inequality and substituting $ \sin x \cos x = \frac{1}{2} \sin(2x)$. This result holds true because the maximum value of $\sin x \cos x$ is 1, and therefore, the inequality can be simplified to $(1+\frac{
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anemone
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For all real numbers a, b, x with $ a, b \geq 0 $ and $ 0 < x < \frac{\pi}{2}$, prove that $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2 \sqrt {ab})^2$.

By using the Cauchy-Schwarz inequality, we can say that
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$ ------(*)

But we know that $ sinxcosx=\frac{sin2x}{2} $
and we're given $ 0 < x < \frac{\pi}{2}$, therefore, $ 0 <sin2x < 1$ and this means $ 0 < \frac{sin2x}{2} < \frac{1}{2}$

From the equation (*), in order to prove $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$, we need to have a maximum of $ sinxcosx $, and this happens when $ sinxcosx=\frac{1}{2} $.

Now, the inequalities becomes $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{\frac{1}{2}})^2$

$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2ab)^2$------(**)

Since $ a, b \geq 0 $,
$ a \geq \sqrt a $
$ b \geq \sqrt b $
$ ab \geq \sqrt {ab} $
$ 2ab \geq 2\sqrt {ab} $
$ 1+2ab \geq 1+2\sqrt {ab} $
$ (1+2ab)^2 \geq (1+2\sqrt {ab})^2 $------(***)

From equations (**) and (***), it's obvious that $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2 \sqrt {ab})^2$.
Am I doing this correct?

Thanks, as usual. :)
 
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  • #2
anemone said:
For all real numbers a, b, x with $ a, b \geq 0 $ and $ 0 < x < \frac{\pi}{2}$, prove that $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2 \sqrt {ab})^2$.

By using the Cauchy-Schwarz inequality, we can say that
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$ ------(*)

But we know that $ sinxcosx=\frac{sin2x}{2} $
and we're given $ 0 < x < \frac{\pi}{2}$, therefore, $ 0 <sin2x < 1$ and this means $ 0 < \frac{sin2x}{2} < \frac{1}{2}$

From the equation (*), in order to prove $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$, we need to have a maximum of $ sinxcosx $, and this happens when $ sinxcosx=\frac{1}{2} $.

Now, the inequalities becomes $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{\frac{1}{2}})^2$

$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2ab)^2$------(**)

Since $ a, b \geq 0 $,
$ a \geq \sqrt a $
$ b \geq \sqrt b $
$ ab \geq \sqrt {ab} $
$ 2ab \geq 2\sqrt {ab} $
$ 1+2ab \geq 1+2\sqrt {ab} $
$ (1+2ab)^2 \geq (1+2\sqrt {ab})^2 $------(***)

From equations (**) and (***), it's obvious that $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+2 \sqrt {ab})^2$.
Am I doing this correct?
The idea of using Cauchy–Schwarz is exactly what is needed here. But you have made mistakes in both the statement and the solution of the problem.

For a start, the problem should say "prove that $\bigl(1+\frac{a}{\sin x}\bigr)\bigl(1+\frac{b}{\cos x}\bigr) \geq (1+ \sqrt {2ab})^2$." As stated, with the 2 outside the square root sign, the result is false (as you can see by putting $a=b=1$ and $x=\pi/4$).

To use Cauchy–Schwarz, you need to apply it to the vectors $\Bigl(1,\sqrt{\frac a{\sin x}}\,\Bigr)$ and $\Bigl(1,\sqrt{\frac b{\cos x}}\,\Bigr).$ The inequality then says $$\Bigl(1+ \tfrac{\sqrt{ab}}{\sqrt{\sin x\cos x}}\Bigr)^2\leqslant \Bigl(1+\frac{a}{\sin x}\Bigr)\Bigl(1+\frac{b}{\cos x}\Bigr).$$

Now you can use the fact that $\sin x\cos x = \frac12\sin(2x)$, as in your argument above, to get the result. That way, you avoid having to assume that $ a \geqslant \sqrt a $ and $ b \geqslant \sqrt b $, which is just as well because those inequalities only hold when $a\geqslant1$ and $b\geqslant1.$
 
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  • #3
Opalg said:
The idea of using Cauchy–Schwarz is exactly what is needed here. But you have made mistakes in both the statement and the solution of the problem.

For a start, the problem should say "prove that $\bigl(1+\frac{a}{\sin x}\bigr)\bigl(1+\frac{b}{\cos x}\bigr) \geq (1+ \sqrt {2ab})^2$." As stated, with the 2 outside the square root sign, the result is false (as you can see by putting $a=b=1$ and $x=\pi/4$).
OK.
I also suspected something went wrong after doing the substitution (a=2, b=3, $x=\pi/6$). But I didn't realize that the 2 should be placed inside the radical sign.

Ha, I couldn't believe I made this kind of mistake! (I've missed out the square root signs on both ab and sinxcosx!:mad:)
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$

After your explanation, everything becomes so clear.:eek:
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{\sqrt {ab}}{\sqrt {sinxcosx}})^2$
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{\sqrt {2ab}}{\sqrt {sin2x}})^2$

But we have $ 0 < sin2x < 1$, to obtain $(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{\sqrt {2ab}}{\sqrt {sin2x}})^2$, we need to have a maximum of $ sinxcosx $, and this happens when $ sinxcosx=1 $.

Therefore,
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{\sqrt {2ab}}{1})^2$
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\sqrt {2ab})^2$ (Q.E.D.)

Thanks, Opalg.
 
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FAQ: Prove that (1+a/sinx)(1+b/cosx) is greater than or equal to (1+2sqrt(ab))^2

1. How can I prove that (1+a/sinx)(1+b/cosx) is greater than or equal to (1+2sqrt(ab))^2?

In order to prove this statement, we can use the fact that for any two positive numbers, a and b, (a+b)^2 is always greater than or equal to 4ab. By substituting a/sinx and b/cosx for a and b respectively, we can see that (1+a/sinx)(1+b/cosx) is equivalent to (1+2sqrt(ab))^2. Therefore, by the transitive property of inequalities, (1+a/sinx)(1+b/cosx) is greater than or equal to (1+2sqrt(ab))^2.

2. Why is it important to prove this statement?

Proving this statement is important because it demonstrates a useful property of trigonometric functions and can be applied to various mathematical problems. It also helps to strengthen our understanding of inequalities and how they can be manipulated to prove different statements.

3. What are the key steps in the proof?

The key steps in the proof are recognizing the equivalence between (1+a/sinx)(1+b/cosx) and (1+2sqrt(ab))^2, using the fact that (a+b)^2 is greater than or equal to 4ab, and applying the transitive property of inequalities to show that (1+a/sinx)(1+b/cosx) is greater than or equal to (1+2sqrt(ab))^2.

4. Can this statement be proven using other methods?

Yes, there are other methods that can be used to prove this statement. One possible method is using algebraic manipulations to expand both sides of the inequality and then using the properties of trigonometric functions to simplify the expressions.

5. Are there any restrictions on the values of a and b in order for this statement to hold?

Yes, in order for this statement to hold, a and b must be positive numbers. This is because we are using the fact that (a+b)^2 is always greater than or equal to 4ab, which only holds true for positive numbers. Additionally, for the original statement to hold, a/sinx and b/cosx must also be positive numbers.

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