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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

math110
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  • using $x=tan(\alpha)$ and $y=tan(\beta)$, I think the function becomes like: $\frac{cos(\alpha+\beta)}{2+\sqrt{2-sin(2\alpha)cos(\beta)-sin(2\beta)cos(\alpha)}}$. Not yet sure what to do next. – tpb261 Jun 09 '14 at 10:15
  • Certainly we have equality if $x=\frac12$ and $y=0$ or conversely. – Bart Michels Jun 09 '14 at 11:01
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    Hmm... Geometrically, this is equivalent to show the in-radius of the triangle with vertices $(0,0)$, $(x,1)$, $(1,y)$ are at most $\frac{\sqrt{5}-1}{4}$. – achille hui Jun 09 '14 at 16:05
  • But How prove this,@achillehui,Thank you – math110 Jun 09 '14 at 16:23
  • @barto Given that the expression is symmetric in $x$ and $y$ (I mean f(x,y)=f(y,x)), so we could use that and find the maximum of the single variable function, but that would same as using Lagrangian multipliers, which as far as I know is disliked on math.se :). But it isn't simple to solve, though. – tpb261 Jun 10 '14 at 03:36
  • Following up on @achillehui's comment, this problem is equivalent to showing that the perimeter of such a triangle is at least $\sqrt5+1$. (This perimeter is achieved with the isosceles triangles where $x=\frac12, y=0$ or $x=0, y=\frac12$.) – Théophile Jun 10 '14 at 04:24
  • @Théophile it's not equivalent. Actually the minimal perimeter is obtained when $x=y=1$, giving $2\sqrt2\approx2.82<1+\sqrt5\approx3.23$. – Bart Michels Jun 10 '14 at 08:10
  • Perhaps it can be proved that the outer radius, that is, $\frac{\sqrt{(1+x^2)(1+y^2)((1-x)^2+(1-y)^2)}}{2-2xy}$ is at most $\frac{\sqrt5-1}2$. – Bart Michels Jun 10 '14 at 08:20
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    We know the area of largest ellipse (the Steiner inellipse) inside a triangle is $\frac{\pi}{3\sqrt{3}}$ of the area of the triangle. This give us an upper bound $\sqrt{\frac{1-xy}{6\sqrt{3}}}$ for RHS. The coefficient $\frac{1}{\sqrt{6\sqrt{3}}}$ is about 0.38% larger than $\frac{\sqrt{5}-1}{4}$. This implies when $xy \gtrsim 0.76%$, the desired inequality is true. We only need to concentrate on what happens when $xy \lesssim 0.76%$. – achille hui Jun 10 '14 at 09:01
  • Is the area to perimeter ratio of a triangle the greatest in the case of an equilateral triangle? If so, since the in-radius is twice this ratio we could examine it in the case of the equilateral triangle with vertices of the form $(0,0)$, $(x,1)$, $(1,y)$. I did this, however my calculations give the in-radius to be $\frac{\sqrt{3} - 1}{\sqrt{6}}$. While this is less than $\frac{\sqrt{5} - 1}{4}$ I was expecting it to be equal. – Ross Pure Jun 10 '14 at 09:13
  • @barto Oh, of course. I think I was up too late past my bedtime... – Théophile Jun 10 '14 at 15:04
  • @RossPure yes, an equilateral triangle maximizes that ratio when the outer radius is fixed. This is not the case here. Besides, an equilateral triangle could never give something with $\sqrt5$, only $\sqrt2,\sqrt3$-ish things. (I did the same calculations as you until I realised this.) – Bart Michels Jun 10 '14 at 15:28
  • Can't this be solved by simply finding the local extrema, or am I just clueless? – Ryan Jun 11 '14 at 04:54
  • http://www.artofproblemsolving.com/Forum/viewtopic.php?t=161686 – nikoma Jun 12 '14 at 09:41

1 Answers1

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

user169301
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