Para números reales positivos $a,b,c,d$ tales que\n$$\n\dfrac{a^2}{b+c+d} + \dfrac{b^2}{a+c+d} + \n\dfrac{c^2}{a+b+d} = \dfrac{3d^2}{a+b+c}\n$$\ndemuestre que\n$$\n\dfrac{3}{a}+ \dfrac{3}{b} + \dfrac{3}{c}+ \dfrac{3}{d} \geq \dfrac{16}{a+b+2d} + \dfrac{16}{b+c+2d} +\n\dfrac{16}{a+c+2d}.\n$$