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

Solution 1:

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 $\sqrt{1+(a+b)^2}+\sqrt{1+(1-a-b)^2} \geq\sqrt{(1+1)^2+(a+b+1-a-b)^2}=\sqrt{5}$