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How to solve the equation: $\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\dfrac{x+y+z}{2}$, where $x$, $y$ and $z$ are reals.

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Hint: The equation is equivalent to \begin{aligned} 0&=x-2\sqrt{x}+y-2\sqrt{y-1}+z-2\sqrt{z-2}\\ &=[x-2\sqrt{x}+1]+[(y-1)-2\sqrt{y-1}+1]+[(z-2)-2\sqrt{z-2}+1]\\ &=(\cdots)^2+(\cdots)^2+(\cdots)^2 \end{aligned}

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We have: $\sqrt{x} = \sqrt{1\cdot x} \le \dfrac{1+x}{2}, \sqrt{y-1} = \sqrt{1(y-1)} \le \dfrac{1+(y-1)}{2}, \sqrt{z-2} = \sqrt{1(z-2)} \le \dfrac{1+(z-2)}{2}$. Add these inequalities, then $LHS \le RHS$, thus $= $ occurs at $1 = x, 1 = y-1 \implies y = 2, 1 = z-2 \implies z = 3$.

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