Given the real values $x=\sqrt{6}-\sqrt{2}$ and $y=\sqrt{5}-\sqrt{3}$, determine the correct inequality relationship between $x$ and $y$.

The correct option is $x>y$.

To determine the correct inequality relationship between the given real numbers $x=\sqrt{6}-\sqrt{2}$ and $y=\sqrt{5}-\sqrt{3}$, we can compare their squares.

First, note that both $x$ and $y$ are positive real numbers because $\sqrt{6}>\sqrt{2}$ and $\sqrt{5}>\sqrt{3}$. For any positive numbers $x$ and $y$, if $x^2>y^2$, then $x>y$.

Step 1: Calculate $x^2$

Using the algebraic identity ${(a-b)}^2=a^2+b^2-2ab$:

$x^2={(\sqrt{6}-\sqrt{2})}^2$

$x^2={\left(\sqrt{6}\right)}^2+{\left(\sqrt{2}\right)}^2-2\left(\sqrt{6}\right)\left(\sqrt{2}\right)$

$x^2=6+2-2\sqrt{12}$

$x^2=8-2\sqrt{12}$

Step 2: Calculate $y^2$

Using the same algebraic identity:

$y^2={(\sqrt{5}-\sqrt{3})}^2$

$y^2={\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\left(\sqrt{5}\right)\left(\sqrt{3}\right)$

$y^2=5+3-2\sqrt{15}$

$y^2=8-2\sqrt{15}$

Step 3: Compare $x^2$ and $y^2$

We compare the two expressions:

$x^2=8-2\sqrt{12}$

$y^2=8-2\sqrt{15}$

Since $12<15$, we have $\sqrt{12}<\sqrt{15}$, which implies:

$2\sqrt{12}<2\sqrt{15}$

Subtracting a smaller positive number ($2\sqrt{12}$) from 8 leaves a larger value than subtracting a larger positive number ($2\sqrt{15}$) from 8. Therefore:

$8-2\sqrt{12}>8-2\sqrt{15}$

$x^2>y^2$

Step 4: Conclusion

Taking the positive square root on both sides yields:

$x>y$