Square root of a positive fraction is greater than itself

1. \(\sqrt{x+y} > 1\)

For the above statement to be true: x+y > 1

You can check with any combination of x and 1-x

0.5, 0.5 : \(\sqrt{0.5} + \sqrt{0.5} => 1.4\)

0.1, 0.9 : \(\sqrt{0.1} + \sqrt{0.9} => 0.31 + 0.94 => 1.25\)

0.01, 0.99 : \(\sqrt{0.01} + \sqrt{0.99} => 0.1 + 0.995 => 1.095\)

It would always be greater than 1

Yes

2. \(x > y > \frac{1}{4}\)

\(\sqrt{1/4} = 0.5\)

So \(\sqrt{x} > \sqrt{y} > 0.5\)

\(\sqrt{x} + \sqrt{y} > 1\)

YES

D.

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