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Let $(x,y) \in \Bbb R^+$ Prove that $$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$

My try

Well, i didn't see a way to factorize this, so i put it in WolframAlpha and i got that we can rewrite it like this, which is obviously true for $(x,y) \in \Bbb R^+$:

$$\frac{(y-x)^2(x+y+1)}{xy(x+y)^2}\ge 0$$

But i don't see the way to get there manually, im stuck here:

$$\frac{(x+y)(y+1)}{xy}\ge \frac{(x+y+2)^2}{(x+y)^2}$$

Michael Rozenberg
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Trobeli
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    At an elementary level the best way to handle inequalities like this is to multiply it all by a common denominator and then expand everything and subtract the RHS from both sides ans then simplify. Then see whether some parts can be re-factored to yield a formula that's never negative. – DanielWainfleet Feb 14 '18 at 20:50

8 Answers8

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Write $a=1/x>0$ and $b=1/y>0$. So we have to prove $$(1+a)(1+b)\geq (1+{2ab\over a+b})^2$$ or $$(a+b)^3+ab (a+b)^2\geq 4ab(a+b)+ 4a^2b^2$$ or $$(a+b)^2(a+b+ab)\geq 4ab(a+b+ ab)$$ or (since $a+b+ab>0$) $$(a+b)^2\geq 4ab$$ which is true.

nonuser
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It is equivalent to $$ \frac{1}{2}\log\left(1+\frac{1}{x}\right)+\frac{1}{2}\log\left(1+\frac{1}{y}\right) \geq \log\left(1+\frac{1}{\frac{x+y}{2}}\right) $$ i.e. to the midpoint-convexity of $f(z)=\log\left(1+\frac{1}{z}\right)$ over $\mathbb{R}^+$. It simply follows from convexity, and convexity is a consequence of $f''(z)=\frac{1}{z^2}-\frac{1}{(z+1)^2}>0$.

Jack D'Aurizio
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It's just Jensen for $f(x)=\ln\left(1+\frac{1}{x}\right)$.

Indeed, $f''(x)=\frac{(2x+1)}{x^2(x+1)^2}>0$ and we get $$\frac{\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{x}\right)}{2}\geq\ln\left(1+\frac{1}{\frac{x+y}{2}}\right),$$ which is your inequality.

Also, it's $$1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\geq1+\frac{4}{x+y}+\frac{4}{(x+y)^2},$$ which is true by C-S $$\frac{1}{x}+\frac{1}{y}\geq\frac{(1+1)^2}{x+y}=\frac{4}{x+y}$$ and AM-GM $$\frac{1}{xy}\geq\frac{1}{\left(\frac{x+y}{2}\right)^2}=\frac{4}{(x+y)^2}.$$

Michael Rozenberg
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it is just $AM-GM$ multiplying out we get $$(x+y)^2+\frac{(x+y)^2}{x}+\frac{(x+y)^2}{y}+\frac{(x+y)^2}{xy}\geq (x+y)^2+4+4(x+y)$$ and this is $$xy(x+y)^2+y(x+y)^2+x(x+y)^2+(x+y)^2\geq xy(x+y)^2+4xy+4xy(x+y)$$ $$(x+y)^3+(x+y)^2\geq 4xy+4xy(x+y)$$ and this is $$(x+y)^2(x+y+1)\geq 4xy(x+y+1)$$

Dr. Sonnhard Graubner
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Why did nobody mention the one-liner solution:

$$\left(1+\frac 1x\right)\left(1+\frac 1y\right)\stackrel{\text{Cauchy-Schwarz}}\geq\left(1+\frac {1}{\sqrt{xy}}\right)^2\stackrel{\text{AM-GM}}\geq\left(1+\frac{1}{\frac{x+y}{2}}\right)^2 = \left(1+\frac{2}{x+y}\right)^2.$$

dezdichado
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Since $\mathbb R^+×\mathbb R^+$ is convex, and the problem is symmetric in $x$ and $y$, the minimum of $$(1+\frac1x)(1+\frac1y)-(1+\frac2{x+y})^2 $$ occurs when $x=y$...

If $x=y$ we get $$(1+\frac1x)(1+\frac1x)-(1+\frac2{2x})^2=0$$...

It's a minimum since for instance $(1,2)$ gives $3-\frac53=\frac43\gt 0$...

  • As I understand it, you can only state that there is a local extremum at $x=y$, but that need not be a global one. Essentially the same point as here: https://math.stackexchange.com/a/2600784/42969. – Martin R Feb 15 '18 at 07:49
  • Yes, i should probably stop using the argument until I can understand it better. I was reading about it here, https://mathoverflow.net/questions/58721/when-does-symmetry-in-an-optimization-problem-imply-that-all-variables-are-equal but honestly still haven't seen the proof all the way through. .. it is an interesting phenomenon though... –  Feb 15 '18 at 08:03
  • Same problem here: https://math.stackexchange.com/a/2647104/42969. – Martin R Feb 15 '18 at 08:04
  • Yes. I was just going to ask you what you thought of that. So you don't think the convexity condition is enough? For instance, i couldn't do this if the region was a sphere (not convex)... it's not a proof, but I was encouraged by the realization that this type of problem would be filtered out, as I came across one such, where it didn't work... –  Feb 15 '18 at 08:21
  • Evidently it's called the Purkiss principle... –  Feb 15 '18 at 08:39
  • Then this would be relevant: https://math.stackexchange.com/a/683668/42969: "The main problem with the Purkiss Principle, most of the time, is to prove that the extreme found is global." – Martin R Feb 15 '18 at 08:41
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Hint:$$\left (\frac{1}{x}-\frac{1}{y}\right )^2\geq 0$$

cansomeonehelpmeout
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I give a proof using Cauchy-Schwartz for fun.

$$\Bigr(1+\frac{1}{x}\Bigl)\Bigr(1+\frac{1}{y}\Bigl)\ge \Bigr(1+\frac{2}{x+y}\Bigl)^2$$ $$\iff \Bigr((x+y)(1+\frac{1}{x})\Bigl)\Bigr((x+y)(1+\frac{1}{y})\Bigl)\ge \Bigr(x+y+2\Bigl)^2$$ $$\iff \Bigr(x+y+1+\frac{y}{x}\Bigl) \Bigr(x+y+1+\frac{x}{y}\Bigl) \ge \Bigr(x+y+2\Bigl)^2$$ $$\iff \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{y}{x}}^2\Bigl) \Bigr(\sqrt x^2+\sqrt y^2+1^2+\sqrt{\frac{x}{y}}^2\Bigl) \\ \ge (\sqrt x \cdot \sqrt x + \sqrt y \cdot \sqrt y + 1\cdot1 + \sqrt{\frac{y}{x}} \cdot \sqrt{\frac{x}{y}})^2 $$

GNUSupporter 8964民主女神 地下教會
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