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# Is this correct?

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5 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
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I think your proof is fine.

Aug 2, 2019
#2
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I did not look at it thoroughly but it seems reasonable to me. Aug 2, 2019
#3
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You have to prove that the parallelogram exists first

Aug 2, 2019
#4
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Thanks!

Aug 2, 2019
#5
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Davis, that proof is incorrect

Guest Aug 3, 2019