<s>[INST] Prove $\sqrt{x-1} + \sqrt{y-1} \le xy, x \ge 1, y \ge 1$ Let $x$ and $y$ be real numbers, such that $x \ge 1$ and $y \ge 1$.
Prove that this inequality is true: $\sqrt{x-1} + \sqrt{y-1} \le xy$
Can someone show me steps to solve it.
PS:I need to give steps on how to solve it.
 [/INST] Using Cauchy Schwarz and AM-GM
$$\sqrt{x-1}+\sqrt{y-1} \le \sqrt{2(x+y-2)} \le \frac{2(x+y-2)+1}{2} \le x+y-1 $$
Now it suffices to prove that
$$x+y-1 \le xy$$
Wich is equivalent to $$(x-1)(y-1) \ge 0$$
So we are done.
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