maximum value of expression $(x-1)(y-1)+(1-\sqrt{1-x^2})(1-\sqrt{1-y^2})$

Question

If $x,y\in\mathbb{R}$. Then maximum value of $$(x-1)(y-1)+(1-\sqrt{1-x^2})(1-\sqrt{1-y^2}).$$

What I try , Here

$1-x^2\geq 0\Longrightarrow x^2\leq 1\Longrightarrow x\in[-1,-1]$ and also $y\in[-1,1]$

So i substitute $x=\sin\alpha$ and $y=\sin\beta$:

$$(\sin\alpha -1)(\sin\beta -1)+(1-\cos\alpha)(1-\cos\beta).$$

$$ \sin\alpha \sin\beta -\sin\alpha-\sin\beta+1+1-\cos\alpha-\cos\beta+\cos\alpha\cos\beta.$$

$$2+\cos(\alpha-\beta)-[\sin\alpha+\sin\beta+\cos\alpha+\cos\beta]$$

I thing that we can solve it using AM GM,

but I did not know. How do I solve it, help me

Answer 1

$$(x-1)(y-1)+\left(1-\sqrt{1-x^2}\right)\left(1-\sqrt{1-y^2}\right)=$$ $$=(x-1)(y-1)+\frac{x^2y^2}{\left(1+\sqrt{1-x^2}\right)\left(1+\sqrt{1-y^2}\right)}\leq1-x-y+xy+x^2y^2\leq5.$$ The equality occurs for $x=y=-1.$

Answer 2

Let $\thinspace f(x)=1-x\thinspace$ and $\thinspace g(x)=1-\sqrt {1-x^2}\thinspace $, where $\thinspace -1≤x≤1\thinspace .$

As you know, you have :

$$ \begin{align}\max \{f(x)\mid -1≤x≤1 \}&=2 \thinspace\thinspace \text{at}\thinspace\thinspace x=-1\end{align} $$

$$ \begin{align}\max \{g(x)\mid -1≤x≤1\}&=1\thinspace\thinspace \text{at}\thinspace\thinspace x=±1\end{align} $$

This implies that :

$$ \begin{align}f(x)f(y)+g(x)g(y)&≤2^2+1^2=5\end{align} $$

The equality holds iff, when $\thinspace (x,y)=(-1,-1)\thinspace .$

Answer 3

You can apply Cauchy-Schwarz inequality twice like this:

\begin{align} &\quad\ \ (1-\sin \alpha)(1 - \sin \beta)+ (1 - \cos \alpha)(1 - \cos \beta)\\ &\le \sqrt{(1 - \sin \alpha)^2+(1 - \cos \alpha)^2}\sqrt{(1 - \sin \beta)^2+(1 - \cos \beta)^2}\\ &=\sqrt{(3 - 2(\sin \alpha +\cos \alpha))(3 - 2(\sin \beta + \cos \beta))} \\ &\le 3+2\sqrt{2} \end{align}

and this turns out to be the max value.

Answer 4

According to your first idea we need to introduce an absolute value for the cosine terms and then, using Cauchy–Schwarz inequality, we obtain

$$(1-\sin\alpha)(1-\sin\beta )+(1-\color{red}{|\cos\alpha|})(1-\color{red}{|\cos\beta|}) \le$$

$$\le \sqrt{(3-2\sin\alpha-2|\cos\alpha|)(3-2\sin\beta-2|\cos\beta|)}\le 5$$

indeed $3-2\sin x-2|\cos x| \le 3-2\cdot (-1)-2\cdot 0=5$ with equality for $x=\frac 3 2 \pi +2k\pi$.