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Triangle Inequality

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The Triangle Inequality says that in a nondegenerate triangle ABC:

AB + BC > AC

BC + AC > AB

AC + AB > BC

That is, the sum of the lengths of any two sides is larger than the length of the third side. In degenerate triangles, the strict inequality must be replaced by "greater than or equal to."


The Triangle Inequality can also be extended to other polygons. The lengths a_1, a_2, \ldots, a_n can only be the sides of a nondegenerate n-gon if a_i < a_1 + \ldots + a_{i -1} + a_{i + 1} + \ldots + a_n = \left(\sum_{j=1}^n a_j\right) - a_i for i = 1, 2 \ldots, n. Expressing the inequality in this form leads to 2a_i < P, where P is the sum of the a_j, or a_i < \frac{P}{2}. Stated in another way, it says that in every polygon, each side must be smaller than the semiperimeter.


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Given a,b,c,d>0, prove:

\sqrt{(a+c)^2+(b+d)^2}+\frac{2|ad-bc|}{\sqrt{(a+c)^2+(b+d)^2}}\geq \sqrt{a^2+b^2}+\sqrt{c^2+d^2} \geq \sqrt{(a+c)^2+(b+d)^2}

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