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

#### anemone

##### MHB POTW Director
Staff member
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
Staff member
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

##### MHB POTW Director
Staff member
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!)
$(1+\frac{a}{sinx})(1+\frac{b}{cosx}) \geq (1+\frac{ab}{sinxcosx})^2$

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

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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