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Is my proof correct? Do I need to make any style changes or is it good?

Let a, b, and c be positive real numbers. Prove that \(\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}.\)


We want to prove that \(\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}\). Consider the above paralelogram where \(AB=c, AC=b, BC=a, DA=a, \angle{DAB}=60^{\circ}\), and \(\angle{DAC}=60^{\circ}\). Applying the Law of Cosines on the triangles DAB, DAC, and ABC gives us the following assuming a, b, and c are side lengths of a triangle.
\(b^2=c^2+a^2-2ac\cos{60^{\circ}}\)
\(c^2=b^2+a^2-2ab\cos{60^{\circ}}\)
\(a^2=b^2+c^2-2bc\cos{120^{\circ}}\)
We can simplify this to get following assuming a, b, and c are side lengths of a triangle.
\(b=\sqrt{c^2+a^2-ac}\)
\(c=\sqrt{b^2+a^2-ab}\)
\(a=\sqrt{b^2+c^2+bc}\)
Using the Triangle Inequality, we can get \(\sqrt{a^2 - ab + b^2}+\sqrt{a^2-ac+c^2}\ge\sqrt{b^2 + bc + c^2}\). Therefore, \(\sqrt{a^2 - ab + b^2} + \sqrt{a^2 - ac + c^2} \ge \sqrt{b^2 + bc + c^2}\) for any positive real numbers a, b, and c.

QED.
 

 Aug 1, 2019
 #1
avatar+1342 
+3

I think your proof is fine.

 Aug 2, 2019
 #2
avatar+103715 
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I did not look at it thoroughly but it seems reasonable to me.   laugh

 Aug 2, 2019
 #3
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+2

You have to prove that the parallelogram exists first

 Aug 2, 2019
 #4
avatar+507 
+1

Thanks!

 Aug 2, 2019
 #5
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+1

Davis, that proof is incorrect

Guest Aug 3, 2019

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