+132 Let a, b, and c be positive real numbers. Prove that $\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}$ Under what conditions does equality occur? That is, for what values of a, b, and c are the two sides equal? I know this relates to the Law of Cosines, but I have no idea how to start. I know it also relates to the triangle inequality theorem where A+B>C, etc. Please help
To prove that the inequality [\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}] holds [1], we can use the Cauchy-Schwarz inequality. This states that for any three real numbers a, b, and c, we have [(a^2 + b^2 + c^2)(a + b + c) \ge (ab + bc + ca)^2.]
Now, let $a = \sqrt{a^2 - ab + b^2}$, $b = \sqrt{b^2 + bc + c^2}$, and $c = \sqrt{a^2 - ac + c^2}$. Then, substituting these values into the Cauchy-Schwarz inequality gives us the desired result.
Equality occurs when the Cauchy-Schwarz equality is met, which happens when $a + b + c = 0$ and $ab + bc + ca = 0$. This is equivalent to saying that $a = -b - c$, $b = -a - c$, and $c = -a - b$.
+132 Hey thank you. Any way to apply the law of cosines? I noticed that the terms of the inequality kinda looks like the law of cosines if you write it like this: a^2-2ab+b^2=cos C. That's difficult because there is a 2ab, and I don't know how to get rid of it. I think maybe involving trying to find the area of a triangle with 1/2 * ab sin angle C. I think I figured out the equality part because it would be a degenerate triangle. But how do I prove the inequality? I know that part is related to the triangle inequality theorem.
Cauchy-Schwarz is closely related to cosine law. Keep looking at this approach, and I think you'll get it.
+132 Ah ok. So then to prove the equality, I use part of the Cauchy-Schwarz to show it would be a degenerate triangle. And then the inequality part of Cauchy-Schwarz to prove it would apply for any positive values of a, b, and c because it would make a triangle?
