Prove Triangle Inequality: $\sqrt{2}\sin A-2\sin B+\sin C=0$

In summary, in a triangle $ABC$ with $\sqrt{2}\sin A-2\sin B+\sin C=0$, it can be proven that $\dfrac{3}{sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$. This holds for any set of angles that satisfy the given equation and the equality holds when $A=\dfrac{\pi}{3}$ and $C=\dfrac{\pi}{4}$.
  • #1
anemone
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In a triangle $ABC$ with $\sqrt{2}\sin A-2\sin B+\sin C=0$, prove that $\dfrac{3}{\sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$.
 
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  • #2
anemone said:
In a triangle $ABC$ with $\sqrt{2}\sin A-2\sin B+\sin C=0$, prove that $\dfrac{3}{\sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$.
I can find _a_ triangle that satisfies $\sqrt{2}\sin A-2\sin B+\sin C=0$ where $\dfrac{3}{sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$ but you use $\ge$ , does this mean there are a multitude of triangles that meet these requirements?

To find my triangle I use the equalities: $\dfrac{sin A}{a} = \dfrac{sin B}{b} = \dfrac{sin C}{c}$

I rearrange your equation to: $\sqrt{2}\sin A+\sin C=2\sin B$ and let $b = 1$ then
$\sqrt{2}\sin A+\sin C=2\dfrac {sin B}{b}$. Now since $\dfrac {sin B}{b} = \dfrac{sin C}{c} = \dfrac{sin A}{a}$
we can write
$\sqrt{2}\sin A+\sin C= \dfrac{sin A}{a} + \dfrac{sin C}{c}$
and if we let $a = \dfrac {1}{\sqrt{2}}$ and $ c = 1$ then our equation is satisfied.

Now that we have found the 3 sides of our triangle we can find the 3 angles. From my CRC handbook I find
$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} = 0.75$
$\cos B = \dfrac{c^2 + a^2 - b^2}{2ca} = 0.3536$
$\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = 0.3536$

Converting to the angles in degrees I get:
$A = 41.41 , B = 69.30 , C = 69.30 $ (as a crosscheck we can see that they sum to 180)
and then
$\dfrac{3}{sin A} + \dfrac{\sqrt{2}}{sin C} = \dfrac{3}{sin 41.41} + \dfrac{\sqrt{2}}{sin69.30} = \dfrac{3}{0.6614} + \dfrac{\sqrt{2}}{0.9354} = 4.5358 + 1.5119 = 6.0477$ and this is greater than $2(\sqrt{3}+1) = 5.4641$.
 
  • #3
DavidCampen said:
I can find _a_ triangle that satisfies $\sqrt{2}\sin A-2\sin B+\sin C=0$ where $\dfrac{3}{sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$ but you use $\ge$ , does this mean there are a multitude of triangles that meet these requirements?

The question says as long as we can prove that the angles in a triangle $ABC$ satisfy $\sqrt{2}\sin A-2\sin B+\sin C=0$, then $\dfrac{3}{\sin A}+\dfrac{\sqrt{2}}{\sin C}\ge 2(\sqrt{3}+1)$ will always hold. (Nod)

The other comment that I have for this unsolved challenge is that the equality holds when $A=\dfrac{\pi}{3}$ and $C=\dfrac{\pi}{4}$.
 

Related to Prove Triangle Inequality: $\sqrt{2}\sin A-2\sin B+\sin C=0$

What is the Triangle Inequality?

The Triangle Inequality states that the sum of any two sides of a triangle must be greater than the third side. Mathematically, it can be written as a + b > c, where a, b, and c are the sides of the triangle.

How do you prove the Triangle Inequality?

To prove the Triangle Inequality, we can use the Pythagorean Theorem and the fact that the sine function is always less than or equal to 1. We can rewrite the given equation as √2sinA + sinC = 2sinB. Then, using the Pythagorean Theorem, we can substitute sinA = √(1-cos²A) and sinC = √(1-cos²C), and simplify to get 2cos²B - 2cosAcosC = 0. Finally, using the fact that cosAcosC ≤ 1/2, we can conclude that 2cos²B - 2cosAcosC ≥ 0, which proves the Triangle Inequality.

What is the significance of proving the Triangle Inequality?

Proving the Triangle Inequality is important in mathematics as it is a fundamental concept in geometry and trigonometry. It helps us understand and analyze triangles, and it is also used in various applications such as engineering, physics, and computer graphics.

Can the Triangle Inequality be applied to all triangles?

Yes, the Triangle Inequality applies to all triangles, regardless of their size or shape. It is a universal rule that must be satisfied for any given triangle.

Are there any other ways to prove the Triangle Inequality?

Yes, there are other ways to prove the Triangle Inequality, such as using the Law of Cosines or the Law of Sines. However, the method described in the given equation is one of the most common and straightforward ways to prove it.

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