A problem on Euclidean Geometry

Question

How can i prove that if a triangle has sides of lengths a, b, e, then its area S satisfies the inequality $$4\sqrt{3}\leq a^{2}+b^{2}+ c^{2}$$

with equality holding only for equilateral triangles. (Hint: If $\theta$ is the angle between sides $b$ and $c$, chosen so that it is at most $60$, then use the formulas $$2S= b c \sin \theta$$ $$2b c \cos \theta= b^{2} + c^{2} - a^{2}$$ $$\cos (60 - \theta) = \dfrac{(\cos \theta)+ \sqrt{3}\sin \theta}{2}$$ Thanks

This is called Weitzenböck's inequality. – user236182 Nov 08 '15 at 16:27 — user236182, Nov 08 '15 at 16:27

score 0 · Accepted Answer · answered Nov 08 '15 at 16:34

Note that

$$\cot{A}=\dfrac{b^2+c^2-a^2}{4S}\implies{}\sum_{\text{cyc}}\cot{A}=\dfrac{a^2+b^2+c^2}{4S}.$$

Hence by Jensen's Inequality for $f(x)=\cot{x}$ we have,

$$\dfrac{a^2+b^2+c^2}{4S}=\sum_{\text{cyc}}\cot{A}\ge{}\sqrt{3}.$$

Equality occurs iff $\angle{A}=\angle{B}=\angle{C}=60\implies{}a=b=c.$

A problem on Euclidean Geometry

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