Inequality with 4 variables

By using AM-GM inequality, we have:

$P=(x+y)(z+t)$

$\quad =\sqrt{(x^2+y^2+2xy)(z^2+t^2+2zt)}$

$\quad=\sqrt{(x^2+y^2)(z^2+t^2)+2xy(z^2+t^2)+2zt(x^2+y^2)+4xyzt}$

$\quad\leq\sqrt{\left(\dfrac{x^2+y^2+z^2+t^2 }{2}\right)^2+2(xz+yt)(xt+yz)+16}$

$\quad\leq\sqrt{25+2\left(\dfrac{xz+yt+xt+yz}{2}\right)^2+16}$

$\quad=\sqrt{41+\dfrac{1}{2}P^2}$

Solving this inequality gives us that $P\leq \sqrt{82}$ .

The equality conditions are:
1. $x^2 + y^2 = z^2 + t^2 = 5$
2. $xyzt = 4$
3. $(x+y)(t+z) = \sqrt{82}$ .

Squaring and expanding condition 3, and substituting in with conditions 1 and 2, we get $( 5 + 2xy)( 5 + \frac{8}{xy} ) = 82$ , or that $xy = \frac{5}{2}, \frac{8}{5}$ .
Thus, solving $x^2+y^2 = 5, xy = \frac{5}{2} \Rightarrow \{ x, y \} = \{ \sqrt{ \frac{5}{2} }, \sqrt{ \frac {5}{2} } \}$ and $z^2 + t^2 = 5, zt = \frac{8}{5} \Rightarrow \{ z, t \} = \{ \frac { \sqrt{ 5 + \frac{16}{5} } + \sqrt{ 5 - \frac{16}{5} } } { 2 } , \frac { \sqrt{ 5 + \frac{16}{5} } + \sqrt{ 5 - \frac{16}{5} } } { 2 } \}$ . Thus, equality can be achieved.

So maximum value of $P$ is $S=\sqrt{82}$ and $\left\lfloor 100S\right\rfloor=\boxed{905}$ .

Note: $z, t$ could be simplitied to $\sqrt{\dfrac{5}{2}\pm\dfrac{3\sqrt{41}}{10}} = \frac{ \sqrt{41} \pm 3 } { \sqrt{20}}$ .

[This is not a complete solution, but it sheds some insight into why the answer isn't just 900.]

For those who think that the answer is 1000, you forgot to account for $x yzt = 4$ , which isn't satisfied when $x = y= z = t = \frac{5}{2}$ .

Most people think that the maximum occurs when $\{ x, y \} = \{ z, t \} = \{ 1, 2 \}$ (in part because of how nice it looks). While this set of values do satisfy the conditions, they do not maximize the product $P$ .

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Note: I used Lagrange multipliers to solve this problem, which has a boundary condition (which is partly what makes this question interesting). You should see Khang's solution for a classical approach.

Instead, let's set $a = x^2 + y^2$ and $b = xy$ . We obtain $10 - a = z^2 + t^2$ and $\frac{4}{b} = zt$ . We then want to find the maximum of

$P^2 = ( x^2 + y^2 + 2xy ) ( z^2 + t^2 + 2zt ) = ( a + 2b ) ( 10 - a + \frac{8}{b} ) .$

We are subject to $x^2 + y^2 + z^2 + t^2 \geq x^2 + y^2 \geq 2 xy \geq 0$ , or that $10 \geq a \geq 2b \geq 0$ and
$x^2 + y^2 + z^2 + t^2 \geq z^2 + t^2 \geq 2 zt \geq 0$ , or that $10 -a \geq \frac{8}{b}$ .

Claim: This is maximized at $\{ (a,b) \} = \{ (5, 2.5) \}$ or $\{ (5, 1.6) \}$ .
I used Lagrange multipliers to show this. We are at a boundary condition.

We have equality when $\left\{ \{ x, y, \}, \{ z, t \} \right\} = \left\{ \{ \sqrt{2.5 } , \sqrt{2.5 } \}, \{ \approx 0.761, \approx 2.1 \} \right\}$ .

The maximum value of $P = \sqrt{ 82 }$ .

First, I let $xy=a$ (noting $a>0$ ) so that $zt=\frac{4}{a}$ , by the equation $xyzt=4$ . Then I let $x+y=r$ . In addition, I applied the simplifying assumption $x>y$ , which can be justified by symmetry. Then I manipulated $x^2+y^2+z^2+t^2=10$ in the following way: $x^2+2xy+y^2+z^2+2zt+t^2 = 10 + 2a + \frac{8}{a}$ $\Rightarrow (z+t)^2=10+2a+\frac{8}{a}-r^2$ $\Rightarrow P^2=r^2(10+2a+\frac{8}{a})-r^4$ Next, I determined the feasible region in the $ar$ -plane. The equation $a=xy$ alone places only the restriction $a>0$ on the value of $a$ . Next, substitute $t=\frac{4}{az}$ in $x^2+y^2+z^2+t^2=10$ to obtain: $10=x^2+y^2+z^2+\frac{16}{a^2z^2}\geq x^2+y^2+\frac{8}{a}=r^2-2a+\frac{8}{a}$ Note that the inequality, $r^2-2a+\frac{8}{a}\leq 10$ , produced is a sharp inequality (consider $z=\frac{2}{\sqrt{a}}$ ). Next, note that, because $x=\frac{r+\sqrt{r^2-4a}}{2}$ and $y=\frac{r-\sqrt{r^2-4a}}{2}$ , we must have $r\geq 2\sqrt{a}$ because the quantity inside the square root operator must be nonnegative. Combining these two findings gives the region of interest in the $ar$ -plane: $2\sqrt{a}\leq r \leq \sqrt{10+2a-\frac{8}{a}}$ Setting the first partial derivates of $P^2 = r^2(10+2a+\frac{8}{a})-r^4$ w.r.t. $r$ and $a$ equal to zero, I determine there are no local extrema in the region of interest. This means that the maximum must lie on the boundary. Therefore, I test the two boundary curves, $r=\sqrt{10+2a-\frac{8}{a}}$ and $r=2\sqrt{a}$ , both with $1\leq a\leq 4$ . On the boundary curve $r=\sqrt{10+2a-\frac{8}{a}}$ , $P^2$ obtains its maximum of $76+\frac{4}{9}$ when $a=\frac{12}{5}$ . Along the other boundary curve $r=2\sqrt{a}$ , $P^2$ obtains a maximum of $82$ when $a=2.5$ . Therefore, the maximum value of $P$ , subject to the constraints, is $\sqrt{82}$ .