Prove $$0\leqq x^{2}+ y^{2}- 2\,x^{2}y^{2}+ 2\,xy\sqrt{1- x^{2}}\sqrt{1- y^{2}}\leqq 1 \tag{1}$$ with $$\left | x \right |,\,\left | y \right |\leqq 1 \tag{2}$$ My normal work $$\begin{cases} \left | x \right |\leqq 1 \Rightarrow -\,1\leqq x\leqq 1 \Rightarrow x\equiv \cos\alpha\,(\,0\leqq \alpha\leqq \pi\,) \\ \left | y \right |\leqq 1 \Rightarrow -\,1\leqq y\leqq 1 \Rightarrow y\equiv \cos\beta\,(\,0\leqq \beta\leqq \pi\,) \\ P\equiv x^{2}+ y^{2}- 2\,x^{2}y^{2}+ 2\,xy\sqrt{1- x^{2}}\sqrt{1- y^{2}} \end{cases}$$ $$\Rightarrow P= \cos^{\,2}\alpha+ \cos^{\,2}\beta- 2\,\cos^{\,2}\alpha\cos^{\,2}\beta+ 2\,\cos\alpha\cos\beta\sqrt{1- \cos^{\,2}\alpha}\sqrt{1- \cos^{\,2}\beta}$$ Otherwise $$\begin{cases} \sqrt{1- \cos^{\,2}\alpha}= \left | \sin\alpha \right |= \sin\alpha\,(\,\because\,0\leqq \alpha\leqq \pi\,) \\ \sqrt{1- \cos^{\,2}\beta}= \left | \sin\beta \right |= \sin\beta\,(\,\because\,0\leqq \alpha\leqq \pi\,) \end{cases}\begin{cases} \cos^{\,2}\alpha= 1- \sin^{\,2}\alpha \\ \cos^{\,2}\beta= 1- \sin^{\,2}\beta \end{cases}$$ $$P= \cos^{\,2}\alpha+ \cos^{\,2}\beta- 2\,\cos^{\,2}\alpha\cos^{\,2}\beta+ 2\,\cos\alpha\cos\beta\sin\alpha\sin\beta$$ The equality condition is $$\alpha+ \beta= \frac{\pi}{2},\,3\,\frac{\pi}{2}$$ I tried to solve the trigonometric equation of $$\begin{cases} \cos^{\,2}\alpha+ \cos^{\,2}\beta- 2\,\cos^{\,2}\alpha\cos^{\,2}\beta+ 2\,\cos\alpha\cos\beta\sin\alpha\sin\beta= 1 \\ \cos^{\,2}\alpha+ \cos^{\,2}\beta- 2\,\cos^{\,2}\alpha\cos^{\,2}\beta+ 2\,\cos\alpha\cos\beta\sin\alpha\sin\beta= 0 \end{cases}$$ but without success. I need to the help! Thanks!
Little idea
How about using vector? $$\begin{cases} \vec{a}= (\,a_{\,1},\,a_{\,2}\,) \\ \vec{b}= (\,b_{\,1},\,b_{\,2}\,) \end{cases}$$ $$\Rightarrow \left [ \,0\leqq (\,\vec{a}\,.\,\vec{b}\,)^{\,2}\leqq \left | \vec{a} \right |^{\,2}\left | \vec{b} \right |^{\,2}= 1\Leftrightarrow 0\leqq (\,a_{\,1}b_{\,1}+ a_{\,2}b_{\,2}\,)^{\,2}\leqq (\,a_{\,1}^{\,2}+ a_{\,2}^{\,2}\,)(\,b_{\,1}^{\,2}+ b_{\,2}^{\,2}\,)= 1\right ]$$ I can't continue!