Let $a,b,c$ denote the sides of a triangle and $[ABC]$ the area of the triangle as usual. $(a)$ If $6[ABC] = 2a^2+bc$, determine $A,B,C$. $(b)$ For all triangles, prove that $3a^2+3b^2 - c^2 \ge 4 \sqrt{3} [ABC]$.
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Tags: geometry, trigonometry, geometry unsolved
Let $a,b,c$ denote the sides of a triangle and $[ABC]$ the area of the triangle as usual. $(a)$ If $6[ABC] = 2a^2+bc$, determine $A,B,C$. $(b)$ For all triangles, prove that $3a^2+3b^2 - c^2 \ge 4 \sqrt{3} [ABC]$.