<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] you can use schwartz inequality directly
we know ;$a_1 b_1+a_2b_2 \leq \sqrt{a_{1}^2+a_{2}^2} \sqrt{b_{1}^2+b_{2}^2}$
just put $a_1 =\sqrt{x-1}$
$a_2 =1$
$b_1 =1$
and
$b_2 =\sqrt{y-1}$
so you will get ;$\sqrt{x-1} .1+1.\sqrt{y-1} \leq \sqrt{x-1+1}\sqrt{y-1+1}=\sqrt{xy}$
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