Problem

Source: Iran 2024 3rd round Algebra exam P1

Tags: algebra



For positive real numbers $a,b,c,d$ such that $$ \dfrac{a^2}{b+c+d} + \dfrac{b^2}{a+c+d} + \dfrac{c^2}{a+b+d} = \dfrac{3d^2}{a+b+c} $$prove that $$ \dfrac{3}{a}+ \dfrac{3}{b} + \dfrac{3}{c}+ \dfrac{3}{d} \geq \dfrac{16}{a+b+2d} + \dfrac{16}{b+c+2d} + \dfrac{16}{a+c+2d}. $$ Proposed by Mojtaba Zare