Bunuel wrote:

If x and y are positive, is \(\sqrt{x}+\sqrt{y}>1\)?

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

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

So here's the Algebraic approach -

Is \(\sqrt{x}+\sqrt{y}>1\), square both sides to get the rephrased question stem as

Is \(x+y+2\sqrt{xy}>1\)

Statement 1: \(\sqrt{x+y}>1\), square both sides to get

\(x+y>1\), Now add \(2\sqrt{xy}\) to both sides to get

\(x+y+2\sqrt{xy}>1+2\sqrt{xy}\), Now as \(x\) & \(y\) are positives so \(2\sqrt{xy}\) will be positive, hence when a positive number is added to \(1\), it will definitely be greater than \(1\)

So we have \(x+y+2\sqrt{xy}>1\).

SufficientStatement 2: \(x>\frac{1}{4}\) take square root of both sides to get

\(\sqrt{x}>\frac{1}{2}\)----------------------(1)

\(y>\frac{1}{4}\), take square root of both sides to get

\(\sqrt{y}>\frac{1}{2}\)----------------------(2), Now add equations (1) & (2) to get

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

SufficientOption

D