<s>[INST] How find this maximum of this $(1-x)(1-y)(10-8x)(10-8y)$ 
let $x,y\in (0,1)$, and such
  $$(1+x)(1+y)=81(1-x)(1-y)$$
  Prove 
  $$(1-x)(1-y)(10-8x)(10-8y)\le\dfrac{9}{16}$$
I ask  $\dfrac{9}{16}$ is best constant?

PS:I don't like Lagrange Multipliers,becasue this is Hight students problem.
My idea:
$$(1-x)(1-y)(10-8x)(10-8y)=\dfrac{1}{64\cdot 81}(8+8x)(8+8y)(10-8x)(10-8y)$$
since
$$(8+8x)(10-8x)\le\dfrac{18^2}{4}=81,(8+8y)(10-8y)\le 81$$
if and only if $$8+8x=10-8x,8+8y=10-8y\Longrightarrow x=y=\dfrac{1}{8}$$
but this not such
$$(1+x)(1+y)=81(1-x)(1-y)$$
so How find this maximum?
 [/INST] We have $$\frac{81}1=\frac{1+xy+(x+y)}{1+xy-(x+y)}$$
Applying Componendo and dividendo,
$$\frac{1+xy}{x+y}=\frac{41}{40}$$
Let $$\frac{1+xy}{41}=\frac{x+y}{40}=u$$
$$\implies(1-x)(1-y)(10-8x)(10-8y)$$
$$=\{1+xy-(x+y)\}\{36+64(1+xy)-80(x+y)\}$$
$$=(41u-40u)\cdot36(1-16u)=\frac{36}{64}\left[1-\left(8u-1\right)^2\right]\le\frac{36}{64}$$
The equality occurs if $8u-1=0$
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