非選第 1 題：
因為 \(x>z\) 且 \(y>z\)，所以 \(x-z>0\) 且 \(y-z>0\)。
利用 \(\displaystyle \left(\sqrt{x-z}\ \sqrt{y-z}-z\right)^2\geq0\)，
展開後得 \(\displaystyle \left(x-z\right)\left(y-z\right)-2z\sqrt{x-z}\sqrt{y-z}+z^2\geq0\)
\(\displaystyle \Rightarrow z^2 +\left(x-z\right)\left(y-z\right)\geq 2z\sqrt{x-z}\sqrt{y-z}\)
（在不等號兩側同時加上 \(z\left(x-z\right)+z\left(y-z\right)\)）
\(\displaystyle \Rightarrow z^2 +\left(x-z\right)\left(y-z\right)+z\left(x-z\right)+z\left(y-z\right)\geq 2z\sqrt{x-z}\sqrt{y-z}+z\left(x-z\right)+z\left(y-z\right)\)
\(\displaystyle \Rightarrow xy\geq \left(\sqrt{z\left(x-z\right)}+\sqrt{z\left(y-z\right)}\right)^2\)
\(\displaystyle \Rightarrow \left(\sqrt{xy}\right)^2\geq \left(\sqrt{z\left(x-z\right)}+\sqrt{z\left(y-z\right)}\right)^2\)
由於 \(\sqrt{xy}\) 與 \(\displaystyle \sqrt{z\left(x-z\right)}+\sqrt{z\left(y-z\right)}\) 皆非負，
得 \(\displaystyle \sqrt{xy}\geq \sqrt{z\left(x-z\right)}+\sqrt{z\left(y-z\right)}\)
