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

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anemone

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Feb 14, 2012
3,712
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. :)
 

Opalg

MHB Oldtimer
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Feb 7, 2012
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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?
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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anemone

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Feb 14, 2012
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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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