parmenides51 wrote: Given $a, b,c \ge 0, a + b + c = 1$, prove that $(a^2 + b^2 + c^2)^2 + 6abc \ge ab + bc + ac$ (I. Voronovich) Use Schur's inequality $a^4+b^4+c^4+abc(a+b+c) \geq a^3 b+a^3 c+b^3 a+b^3 c+c^3 a+c^3 b$

By computer,we have: $$\text{LHS-RHS}=1/2\, \left( a-b \right) ^{2} \left( -c+a+b \right) ^{2}+1/2\, \left( b-c \right) ^{2} \left( -a+b+c \right) ^{2}+1/2\, \left( c-a \right) ^ {2} \left( -b+c+a \right) ^{2} \geqq 0$$