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Who can help with the following inequality? I can prove it but using some rather ugly approach (e.g. by leveraging the derivative of $\frac{1}{\sqrt{t+1}}+\frac{1}{2}\sqrt{1-\frac{8}{t^2+8}}$ to show this is always less than 1 for $t>0$.

I'm just wondering if we can have some elegant simple prove. I guess we should use Jensen's inequality. Thanks.

Suppose $x,y,z\in R^+$ and $xyz=8$, try to prove that $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}<2.$$

Please note that the usual AM-GM inequality may not do its trick here as the equality is rather hold on the boundary.

Thanks.

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    $\begingroup$ I am assuming that $x, y, z$ are all positive? $\endgroup$ – Prism Jul 18 '13 at 20:57
  • $\begingroup$ Yes. Sorry. Let me amend it. $\endgroup$ – user85356 Jul 18 '13 at 21:01
  • $\begingroup$ We might be able to use a slightly different inequality here: since $\|x\|_1\leq \sqrt n \|x\|_2$ (in an $n$ dimensional space), we have $$\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}}\leq 3 \sqrt{\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}}$$ $\endgroup$ – Omnomnomnom Jul 18 '13 at 21:28
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    $\begingroup$ Oh wow, I'm amazed that $xyz\geq 8$ needs to be used. The 'equality' condition seems to be as $y, z \rightarrow 0$. $\endgroup$ – Calvin Lin Jul 18 '13 at 21:57
  • $\begingroup$ As the LHS is convex, it cannot achieve maximum in the interior, i.e we need one or more of $x=0, y=0$ or $z=0$ satisfied. WLOG $x \le y \le z$, and taking $x \to 0$, we get $z = \frac8{xy} \to \infty$ and LHS $\to 1+\sqrt{\dfrac1{1+y}}+0< 2$, so this is the supremum. The "$8$" as you may see, is purely incidental, it could have been any other number for the same "maximum". $\endgroup$ – Macavity Aug 29 '14 at 6:51
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Let $f(x,y,z)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ and WOLOG assume that $x\geq y\geq z$ so we have $x\geq 2$. Consider the following two cases:

  1. $yz\geq 1$ : In this case we have $z\geq1/y$ and thus $$f(x,y,z)\leq f(x,y,1/y)\\ \leq\sqrt{3\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{y}{1+y}\right)}\\ = \sqrt{3\left(\frac{1}{1+x}+1\right)}\\ \leq \sqrt{3\left(\frac{1}{3}+1\right)}=2.$$ Equation in this case cannot be attained as we cannot have $z=1/y$ and $x=2$ simultaneously.
  2. $yz\leq 1$: Fix the product $yz=p^2$. Since we assumed $y\geq z$ we must have $y\geq p$. Furthermore, we have $p\leq 1$ for which the function $\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}$ would be decreasing with respect to $y$. Therefore, the mentioned function is minimized at $y=z=p$, i.e., $$\frac{1}{\sqrt{1+y}}+\frac{1}{\sqrt{1+z}}\leq\frac{2}{\sqrt{1+\sqrt{yz}}}\\=\frac{2}{\sqrt{1+\sqrt{8/x}}}.$$ Thus $f(x,y,z)\leq \frac{1}{\sqrt{1+x}}+\frac{2}{\sqrt{1+\sqrt{8/x}}}.$ The RHS is increasing in $x$ so the upper bound as $x\to \infty$ would be $f(x,y,z)<2$.

There might be neater way to show the second case.

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Here's a solution using Jensen's inequality.

To get rid of the constraint, let's set $$ x = 2b/a \quad y = 2c/b \quad z = 2a/c $$ the inequality becomes $$ \sqrt{\frac a {a + 2b}} + \sqrt{\frac b {b + 2c}} + \sqrt{\frac c {c + 2a}} < 2 $$ Since $\sqrt t$ is a concave function, applying Jensen's inequality we get $$ \begin{align} \sum_{cyc}\sqrt{\frac a {a + 2b}} &= \sum_{cyc} \frac {a + c} {2(a + b + c)} \sqrt{\frac{4(a + b + c)^2}{(a + c)^2} \frac a {a + 2b}}\\ &\leq \sqrt{\sum_{cyc} \frac{2a(a + b + c)}{(a + c)(a + 2b)}} \end{align} $$ Now, it suffices to expand the expression $E$ inside the last square root.

If we call $N$ its numerator and $D$ its denominator then $4D - N$ results in a sum of monomials in $a, b$ and $c$ with positive coefficients. Therefore $$ E = \frac N D < 4 $$ and the inequality is proved.

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  • $\begingroup$ How did you sort out $4D-N$ results in postive weights? I verified it in Matlab and indeed that is true yet just wondering if there is any trick revealing this more easily. $\endgroup$ – user85356 Jul 20 '13 at 14:22
  • $\begingroup$ @user85356 A full expanding here it's $10$ minutes of work without any software. $\endgroup$ – Michael Rozenberg Mar 18 '17 at 8:18
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$a=\sqrt{x+1}>1,b=\sqrt{y+1}>1, c=\sqrt{z+1}>1$,for $a,b,c$,at least two of them $>2$ or $\le 2$,WOLG, assume $a,b$ either both $>2$ or both $\le2$, $z=\dfrac{8}{xy}=\dfrac{8}{(a^2-1)(b^2-1)}$,

edit:

case I: if $a>2$ and $b>2$

it is trivial $\dfrac{1}{a}+\dfrac{1}{b} <1 ,\dfrac{1}{c}<1$

so it is true in this case.

case II: when $a \le 2,b \le 2$,

now we need to prove:

$\dfrac{1}{a}+\dfrac{1}{b}+\sqrt{\dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}}<2 \iff \dfrac{(a^2-1)(b^2-1)}{(a^2-1)(b^2-1)+8}< \left(2-\dfrac{1}{a}-\dfrac{1}{b} \right)^2 \iff \dfrac{(2ab-a-b)^2}{a^2b^2} > \dfrac{(a^2-1)(b^2-1)}{((a^2-1)(b^2-1)+8)}$

note: $(a^2-1)(b^2-1)+8>a^2b^2 \iff 9>a^2+b^2 \iff (a^2\le 4) \cap (b^2 \le 4) $

now we need to prove $(2ab-a-b)^2>(a^2-1)(b^2-1)$

$(2ab-a-b)^2= (a(b-1)+b(a-1))^2 \ge 4ab(a-1)(b-1) \iff 4ab >(a+1)(b+1) \iff (2a>a+1 )\cap (2b>b+1) \iff (a>1) \cap (b>1)$

it is true.

both cases are true. QED

Indeed, when $xyz=2$, this inequality is also true.

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The problem can be restated as follows. Find the maximum of the function

$$ f(x,y,z)=\sqrt{\frac{1}{1+x}}+\sqrt{\frac{1}{1+y}}+\sqrt{\frac{1}{1+z}} $$

constrained to $xyz-8=0$. You can use a Lagrange multiplier and proceed.

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    $\begingroup$ Yes. That can prove. Can we have some more elementary approach? Maybe taking use of the Jensen's inequality? $\endgroup$ – user85356 Jul 19 '13 at 9:22
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    $\begingroup$ Jensen's inequality doesn't work. $\endgroup$ – chenbai Jul 19 '13 at 11:48

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