JBMO2020 wrote: Let $a, b, c$ be non-negative real numbers. Prove that $$a\sqrt{3a^2+6b^2}+b\sqrt{3b^2+6c^2}+c\sqrt{3c^2+6a^2}=>(a+b+c)^2$$ $$\sum_{cyc}a\sqrt{3a^2+6b^2}-(a+b+c)^2=\sum_{cyc}(a\sqrt{3a^2+6b^2}-a^2-2ab)=$$$$=\sum_{cyc}\frac{2a(a-b)^2}{\sqrt{3a^2+6b^2}+a+2b}\geq0.$$

JBMO2020 wrote: Let $a, b, c$ be non-negative real numbers. Prove that $$a\sqrt{3a^2+6b^2}+b\sqrt{3b^2+6c^2}+c\sqrt{3c^2+6a^2}\ge(a+b+c)^2$$ $$a\sqrt{3a^2+6b^2}\ge a^2+2ab\iff (a-b)^2\ge 0 $$$$a\sqrt{3a^2+6b^2}+b\sqrt{3b^2+6c^2}+c\sqrt{3c^2+6a^2}\ge a^2+2ab+b^2+2bc+c^2+2ca=(a+b+c)^2$$https://artofproblemsolving.com/community/c6h1248991p6422584