Prove Triangle Sides Inequality $\sqrt{b+c-a} \dots \le 3$

[anemone]

Let $a,\,b,\,c$ be the sides of a triangle. Prove that

$\dfrac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\dfrac{\sqrt{c+a-b}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a+b-c}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$.

 

[jvicens]

Thank you for your interesting post. I am always eager to explore and analyze mathematical concepts. After some investigation, I have found that this inequality is known as the Nesbitt's inequality. It is a well-known result in the field of mathematics, specifically in the area of geometric inequalities.

To prove this inequality, we will use the fact that in a triangle, the sum of any two sides is always greater than the third side. This is known as the triangle inequality. Now, let us consider the first term in the given expression, $\dfrac{\sqrt{b+c-a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}$. By using the triangle inequality, we can write the numerator as $\sqrt{b+c-a}=\sqrt{b}+\sqrt{c}-\sqrt{a}-2\sqrt{bc}$. Similarly, we can express the other two terms in the same way.

Substituting these expressions into the original inequality, we get:

$\dfrac{\sqrt{b}+\sqrt{c}-\sqrt{a}-2\sqrt{bc}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\dfrac{\sqrt{c}+\sqrt{a}-\sqrt{b}-2\sqrt{ca}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\dfrac{\sqrt{a}+\sqrt{b}-\sqrt{c}-2\sqrt{ab}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$

Simplifying this expression, we get:

$1-\dfrac{2\sqrt{bc}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+1-\dfrac{2\sqrt{ca}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+1-\dfrac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}-\sqrt{c}}\le 3$

Or, $\dfrac{2\sqrt{bc}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}+\dfrac{2\sqrt{ca}}{\sqrt{c}+\sqrt{a}-\sqrt{b}}+\dfrac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}-