Prove this inequality $a \sqrt{1-b^2}+b\sqrt{1-a^2}\le1 $ Prove that for $a,b\in [-1,1]$:

$$a\sqrt{1-b^2}+b\sqrt{1-a^2}\leq 1$$

 A: Method 1:
HINT:
As $1-b^2\ge 0\implies -1\le b\le 1,$ let $ b=\sin B$
Similarly, $a=\sin A$
$\implies a\sqrt{1-b^2}+b\sqrt{1-a^2}=\sin A\cos B+\cos A\sin B=\sin(A+B)$

Method 2:
Let $a\sqrt{1-b^2}+b\sqrt{1-a^2}=y$
Squaring we get, $y^2=a^2(1-b^2)+b^2(1-a^2)+2ab\sqrt{(1-a^2)(1-b^2)}$
So, $1-y^2$
$=(1-a^2)(1-b^2)+(ab)^2-2ab\sqrt{(1-a^2)(1-b^2)}$
$=\left(\sqrt{(1-a^2)(1-b^2)}-ab\right)^2\ge0$ for real $a,b$
$\implies y^2\le 1\implies  y\le1$

Method 3:
$$\left(a\sqrt{1-b^2}\pm b\sqrt{1-a^2}\right)^2+\left(a\cdot b\mp \sqrt{(1-a^2)(1-b^2)}\right)^2$$
$$=a^2(1-b^2)+b^2(1-a^2)\pm2ab\sqrt{(1-a^2)(1-b^2)}+a^2b^2+(1-a^2)(1-b^2)\mp2ab\sqrt{(1-a^2)(1-b^2)}=1$$
Now, if $p^2+q^2=1$ where $p,q$ are real, $q^2\ge0\implies p^2=1-q^2\le1$
So, each of $a\sqrt{1-b^2}\pm b\sqrt{1-a^2},a\cdot b\mp \sqrt{(1-a^2)(1-b^2)}$ is $\le1$
A: Use the AM-GM inequality: For nonnegative $x,y$, we have $\displaystyle\sqrt{xy}\le\frac{x+y}2$. This follows immediately from expanding and rearranging the obvious inequality $(\sqrt x-\sqrt y)^2\ge0$.
Here, we have $$\begin{array}{rl} a\sqrt{1-b^2}+b\sqrt{1-a^2}&\le|a|\sqrt{1-b^2}+|b|\sqrt{1-a^2}\\&\le\frac{a^2+(1-b^2)}2+\frac{b^2+(1-a^2)}2\\&=\frac22=1.\end{array}$$
Note that the AM-GM inequality is an equality iff $x=y$, so here we have $a=|a|$, $b=|b|$, and $a^2=1-b^2$, or $a^2+b^2=1$, so $a=\sin\theta$ and $b=\cos\theta$ for some $\theta$ in the first quadrant. This suggests the other solution, where more generally we let $a=\sin\alpha$, $b=\cos\beta$ for some $\alpha,\beta$ (or some variant thereof).
A: By Cauchy-Schwarz
$$(a\sqrt{1-b^2}+b\sqrt{1-a^2})^2 \leq (a^2+b^2)(1-b^2+1-a^2)=2(a^2+b^2)-(a^2+b^2)^2$$
$$=1-(1-a^2-b^2)^2\leq 1$$
