Show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$

Question

How can one show that $\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$

Assuming that :

$\sqrt{x-1}+\sqrt{y-1}\leq \sqrt{xy}$

So

$(\sqrt{x-1}+\sqrt{y-1})^2\leq xy$

$\sqrt{(x-1)(y-1)} \leq xy-x-y+2$

$ (y-1)(x-1)+3 \leq \sqrt{(x-1)(y-1)}$

Here I'm stuck !

Answer 1

There are some errors in your calculation, e.g. a missing factor 2 in $$ (\sqrt{x-1}+\sqrt{y-1})^2 = x - 1 + y - 1 + 2\sqrt{x-1}\sqrt{y-1} $$ and in the last step the inequality sign is in the wrong direction and the number $3$ is wrong.

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For $x \ge 1$, $y \ge 1$ you can square the inequality (since both sides are non-negative): $$ \sqrt{\mathstrut x-1}+\sqrt{\mathstrut y-1}\leq \sqrt{\mathstrut xy} \\ \Longleftrightarrow (x-1) + (y-1) + 2 \sqrt{\mathstrut x-1}\sqrt{\mathstrut y-1} \le xy \\ \Longleftrightarrow 0 \le xy - x - y + 2 - 2 \sqrt{\mathstrut (x-1)(y-1)} \\ \Longleftrightarrow 0 \le (x-1)(y-1) - 2 \sqrt{\mathstrut (x-1)(y-1)} + 1 $$ With $t := \sqrt{(x-1)(y-1)}$ the right-hand side is $$ t^2 - 2 t + 1 = (t-1)^2 \ge 0 \, . $$ so that the inequality is true.

It follows also that equality holds if and only if $t = 1$, i.e. if $(x-1)(y-1) = 1$.

Answer 2

Starting with “Assuming that $X$” when $X$ is the thing to be proved is not the right thing to do. You can make your computations easier if you set $$ t=\sqrt{x-1},\quad u=\sqrt{y-1} $$ so $x=t^2+1$ and $y=u^2+1$. The inequality to be proved becomes $$ t+u\le\sqrt{(t^2+1)(u^2+1)} $$ that has a single radical. Since $t+u\ge0$, the inequality is equivalent to $$ (t+u)^2\le(t^2+1)(u^2+1) $$ that is, $$ t^2+2tu+u^2\le t^2u^2+t^2+u^2+1 $$ and so equivalent to $$ 0\le t^2u^2-2tu+1 $$ (by transporting terms to the right-hand side) which becomes $$ 0\le(tu-1)^2 $$ which is true.

Don't forget the $2$ when you square! You basically did $(a+b)^2=a^2+ab+b^2$ and this is where you got stuck.