<s>[INST] How prove $\frac{1-xy}{\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}}\le\frac{\sqrt{5}-1}{4}$ Question:

let $x,y\in [0,1]$, show that
  $$\dfrac{1-xy}{\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}}\le\dfrac{\sqrt{5}-1}{4}$$
  Thank you (I think this inequality can use Geometric interpretation)

my idea:
$$\Longleftrightarrow 4(1-xy)\le (\sqrt{5}-1)[\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}]$$
$$\Longleftrightarrow (\sqrt{5}-1)[\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{(1-x)^2+(1-y)^2}]+4xy\ge 4$$
then I can't prove it.
Thank you
 [/INST] The in equality becomes $\sqrt{1+a^2}+\sqrt{1+b^2}+\sqrt{(1-a)^2+(1-b)^2}\geq (1+\sqrt{5})(1-ab)$
We'll prove that
$\sqrt{(1-a)^2+(1-b)^2}+\sqrt{5}ab\geq \sqrt{1+(1-a-b)^2}$  (1)
$\sqrt{1+a^2}+\sqrt{1+b^2}+ab\geq 1+\sqrt{1+(a+b)^2}$  (2)
To prove (1) you just need to square it
To prove (2), set :
A=$1+\sqrt{1+(a+b)^2}+\sqrt{1+a^2}+\sqrt{1+b^2}$
B=$\sqrt{1+(a+b)^2}+\sqrt{(1+a^2)(1+b^2)}$
We have :
$1+\sqrt{1+(a+b)^2}-\sqrt{1+a^2}-\sqrt{1+b^2}
=\frac{2ab+2[\sqrt{1+(a+b)^2}-\sqrt{(1+a^2)(1+b^2)}]}{A}
=\frac{2ab+\frac{2ab(2-ab)}{B}}{A}\le  \frac{2ab+\frac{4ab}{B}}{A}$
So we only need to prove :
$\frac{1}{A}\geq\frac{2}{B}+1$
This is right because $A\geq4$ and $B\geq2$
From (1) and (2), the inequation become :
$\sqrt{1+(a+b)^2}+\sqrt{1+(1-a-b)^2} \geq\sqrt{5}$
This is right due to minkowski inequality
$\sqrt{1+(a+b)^2}+\sqrt{1+(1-a-b)^2} \geq\sqrt{(1+1)^2+(a+b+1-a-b)^2}=\sqrt{5}$
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