gaokao 2019 Q23

gaokao · China · national-III-science 10 marks Inequalities Optimization Subject to an Algebraic Constraint

Let $x, y, z \in \mathbf{R}$ and $x + y + z = 1$.
(1) Find the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$;
(2) If $(x-2)^2 + (y-1)^2 + (z-a)^2 \geqslant \frac{1}{3}$ holds for all $x, y, z$ satisfying $x + y + z = 1$, prove that $a \leqslant -3$ or $a \geqslant -1$.

(1) Since $[(x-1) + (y+1) + (z+1)]^2 = (x-1)^2 + (y+1)^2 + (z+1)^2 + 2[(x-1)(y+1) + (y+1)(z+1) + (z+1)(x-1)] \leqslant 3[(x-1)^2 + (y+1)^2 + (z+1)^2]$, from the given condition $(x-1)+(y+1)+(z+1) = x+y+z+1 = 2$, we have $(x-1)^2 + (y+1)^2 + (z+1)^2 \geqslant \frac{4}{3}$, with equality if and only if $x = \frac{5}{3}, y = -\frac{1}{3}, z = -\frac{1}{3}$. Therefore, the minimum value is $\frac{4}{3}$.
(2) Since $[(x-2) + (y-1) + (z-a)]^2 \leqslant 3[(x-2)^2 + (y-1)^2 + (z-a)^2]$, and $(x-2)+(y-1)+(z-a) = x+y+z - (3+a) = -(2+a)$, the minimum value of $(x-2)^2 + (y-1)^2 + (z-a)^2$ is $\frac{(2+a)^2}{3}$. From the given condition, $\frac{(2+a)^2}{3} \geqslant \frac{1}{3}$, which gives $(2+a)^2 \geqslant 1$, so $a \leqslant -3$ or $a \geqslant -1$.

Let $x, y, z \in \mathbf{R}$ and $x + y + z = 1$.

(1) Find the minimum value of $(x-1)^2 + (y+1)^2 + (z+1)^2$;

(2) If $(x-2)^2 + (y-1)^2 + (z-a)^2 \geqslant \frac{1}{3}$ holds for all $x, y, z$ satisfying $x + y + z = 1$, prove that $a \leqslant -3$ or $a \geqslant -1$.

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