FelixD wrote: Let $ x$, $ y$, $ z$ be real numbers satisfying $ x^2 + y^2 + z^2 + 9 = 4(x + y + z)$. Prove that \[ x^4 + y^4 + z^4 + 16(x^2 + y^2 + z^2) \ge 8(x^3 + y^3 + z^3) + 27\] and determine when equality holds. Notice that $ x^4-8x^3+16x^2-9+6(x^2-4x+3)=(x-1)^2(x-3)^ 2\ge 0.$ Therefore $ x^4-8x^3+16x^2-9 \ge -6(x^2-4x+3).$ Adding this and its analogous inequalities, we get $ \sum x^4-8\sum x^3+16\sum x^2-27 \ge -6\left( \sum x^2-4\sum x +9\right) =0.$ This proves the desired inequality.

what is MEMO?

A quick search would help you, but anyway: MEMO = Middle European Mathematical Olympiad.

\[ x^{2}+y^{2}+z^{2}+9=4(x+y+z)\] \[ x^{2}-4x+y^{2}-4y+z^{2}-4z+9=0\] \[ (x^{2}-4x+4)+(y^{2}-4y+4)+(z^{2}-4z+4)=3\] \[ (x-2)^{2}+(y-2)^{2}+(z-2)^{2}=3\] \[ x^{4}+y^{4}+z^{4}+16(x^{2}+y^{2}+z^{2})\ge8(x^{3}+y^{3}+z^{3})+27\] \[ (x^{4}-8x^{3}+16x^{2})+(y^{4}-8y^{3}+16y^{2})+(z^{4}-8z^{3}+16z^{2})\ge27\] \[ (x^{2}-4x)^{2}+(y^{2}-4y)^{2}+(z^{2}-4z)^{2}\ge27\] \[ ((x-2)^{2}-4)^{2}+((y-2)^{2}-4)^{2}+((z-2)^{2}-4)^{2}\ge27\] After substition $ a=(x-2)^{2}, b=(y-2)^{2}, c=(z-2)^{2}$ the following inequality equivalently to $ a+b+c=3$ then \[ (a-4)^{2}+(b-4)^{2}+(z-4)^{2}\ge27\] \[ a^{2}+b^{2}+c^{2}-8(a+b+c)+48\ge27\] \[ a^{2}+b^{2}+c^{2}-24+48\ge27\] \[ a^{2}+b^{2}+c^{2}\ge3\] By Cauchy-Schwarz inequality \[ (a^{2}+b^{2}+c^{2})(1+1+1)\ge(a+b+c)^{2}=9\] \[ 3(a^{2}+b^{2}+c^{2})\ge9\] \[ a^{2}+b^{2}+c^{2}\ge3\] Done!