Anyone know ow to do this 4u proofs question? (1 Viewer)

Afiguana · Mar 23, 2024

Does anyone know how to do this?

Lith_30 · Mar 23, 2024

$\frac{b+h}{2x}\geq\frac{\sqrt{bh}}{x}$ assuming that x does not equal 0

$\frac{b}{2x}+\frac{h}{2x}\geq\sqrt{\frac{b}{x}\frac{h}{x}}$

Now from the triangle diagram using trig we know $\sin\left(\frac{\theta}{2}\right)=\frac{\frac{b}{2}}{x}$ and $\cos\left(\frac{\theta}{2}\right)=\frac{h}{x}$ .
Then we can say that $\sin\left(\frac{\theta}{2}\right)+\frac{1}{2}\cos\left(\frac{\theta}{2}\right)\geq\sqrt{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}$
then,
$\sin\left(\frac{\theta}{2}\right)+\frac{1}{2}\cos\left(\frac{\theta}{2}\right)\geq\sqrt{\sin\left(\theta\right)$ .

So $\sin\left(\frac{\theta}{2}\right)+\frac{1}{2}\cos\left(\frac{\theta}{2}\right)\geq\sin\left(\theta\right)$ , cause squaring a positive number less than 1 decreases its value.

Luukas.2 · Mar 23, 2024

Right-angle triangle trigonometry should allow you to show that

$\text{LHS} = \sin{\frac{\theta}{2}} + \frac{1}{2}\cos{\frac{\theta}{2}} = \frac{b + h}{2x}$

Applying the AM-GM inequality leads to

$\text{LHS} \geq \sqrt{\frac{bh}{x^2}}$

Now, using the fact that

$\text{Area of } \Delta = \frac{1}{2}bh = \frac{1}{2}x^2\sin{\theta} \qquad \implies \qquad \frac{bh}{x^2} = \sin\theta$

which leads you to

$\sin{\frac{\theta}{2}} + \frac{1}{2}\cos{\frac{\theta}{2}} \ge \sqrt{\sin\theta} \ge \sin\theta$

as $\theta \in (0,\,2\pi) \implies \sin\theta \in (0,\,1]$ and thus $\sqrt{\sin\theta} \ge \sin\theta.$

Afiguana · Mar 23, 2024

Ah ok thank you so much that makes sense now!

Anyone know ow to do this 4u proofs question? (1 Viewer)

Afiguana

New Member

Attachments

Lith_30

o_o

Luukas.2

Well-Known Member

Afiguana

New Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)