It is known that \(a > 1\). Is it always true that \(\lfloor \sqrt{\lfloor \sqrt{a}\rfloor }\rfloor = \lfloor \sqrt{4}{a}\rfloor\)?
It is known that \(a = x+y + \sqrt{xy}\), \(b = y + z + \sqrt{yz}\), \(c = x + z + \sqrt{xz}\). where \(x > 0\), \(y > 0\), \(z > 0\). Prove that \(a + b + \sqrt{ab} > c\).