$X^3 + Ky \geq (x+1)^3 \Rightarrow Ky \geq 3x^2 + 3x + 1$. Sum cyclically: $K(x+y+z) \geq 3(x^2 + y^2 +z^2) + 3( x+y+z) +3 \Rightarrow K(x+y+z) \geq 6( x+y+z) +3 \Rightarrow $ $K(x+y+z) \geq 7( x+y+z) \Rightarrow K \geq 7$. Check that $x = y = z = 1, K= 7$ works.

My solution at the test.I realize that many people I know use a kind of bash bounding method. Anyway, we have $$\sum{x^3+Ky \ge (x+1)^3}=\sum{x^3+3x^2+3x+1}$$Which is equivalent to, $$(K-3)(x+y+z)\ge\sum{3x^2+1}$$From here, notice that we have $$\sum{(3x-1)(x-1)}\ge0$$$$\sum{3x^2+1}\ge4(x+y+z)$$Therefore, $$(K-3)(x+y+z)\ge4(x+y+z)$$$$K\ge7$$Which proves the $(a) $ part, for the $(b) $part we just need to check the equality case when $x=y=z=1$.

Doppel wrote: My solution at the test.I realize that many people I know use a kind of bash bounding method. Anyway, we have $$\sum{x^3+Ky \ge (x+1)^3}=\sum{x^3+3x^2+3x+1}$$Which is equivalent to, $$(K-3)(x+y+z)\ge\sum{3x^2+1}$$From here, notice that we have $$\sum{(3x-1)(x-1)}\ge0$$$$\sum{3x^2+1}\ge4(x+y+z)$$Therefore, $$(K-3)(x+y+z)\ge4(x+y+z)$$$$K\ge7$$Which proves the $(a) $ part, for the $(b) $part we just need to check the equality case when $x=y=z=1$. yeah i used almost the same approach, i used the fact that $x \ge 1 \implies x^2 \ge x$ to get \[\sum x^3 + Ky \ge (x+1)^3 = x^3 + 3x^2 + 3x + 1 \ge x^3 + 6x + 1 > x^3+6x\]sum cyclically to get \[x^3+y^3+z^3+K(x+y+z) > x^3+y^3+z^3 + 6(x+y+z) \implies K>6\]Hence if $K$ satisfies, then $K \ge 7$. Check that $(x,y,z,K)=(1,1,1,7)$ works, so $K=7$ is the minimum BTW, what province are you from?

Notice that $(x-y)+(y-z)+(z-x)=0$ then $(x-y),(y-z),(z-x)$ must have at least one non-negative number. WLOG, we assume that ${x-y}\ge{0} \Rightarrow {x}\ge{y}$ ${x}^3+Ky$ is cubed $\Rightarrow {{x}^3+Ky}\ge({x+1}^3)={x}^3+3{x}^2+3x+1$ $\Rightarrow {K}\ge{\frac{3{x}^2+3x+1}{y}}>{\frac{3{y}^2+3y}{y}}\ge{6}$ $\Rightarrow {K}\ge{7}$ With$K=7$, we can take $x=y=z=1$.