Given $x,y,z\geq0$ and $xy+yz+zx=1$. Show that $\displaystyle\frac 1 {\sqrt{x+y}}+\frac 1 {\sqrt{y+z}}+\frac 1 {\sqrt{z+x}}\geq2+\frac 1 {\sqrt2}$.

I've tried many things but all failed. The only thing I know is that the equality holds when $x=y=1, z=0$.

Please help. Thank you.

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    $\begingroup$ I think there would be $x=y=1, z=0$ for equality. $\endgroup$ – Mher Jun 7 '13 at 9:33
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    $\begingroup$ There is such identity. $cot(a)cot(b)+cot(b)cot(c)+cot(c)cot(a),$ if $a+b+c=\pi.$ Try substitution $x=a, y=b, z=c.$ $\endgroup$ – Mher Jun 7 '13 at 9:54
  • $\begingroup$ I'm not good at these things, but maybe it's an application of Muirhead's inequality? $\endgroup$ – Daniel McLaury Jun 7 '13 at 11:01

Without loss of generality, we assume that $x\ge y\ge z$.

Lemma 1: $$\dfrac{1}{\sqrt{y^2+1}}+\dfrac{1}{\sqrt{z^2+1}}\ge 1+\dfrac{1}{\sqrt{(y+z)^2+1}}$$

Proof: this inequality is equivalent to $$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}+\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge 1+\dfrac{1}{(y+z)^2+1}+\dfrac{2}{\sqrt{(y+z)^2+1}}.$$ Notice that $$(y+z)^2+1-(y^2+1)(z^2+1)=yz(2-yz)\ge 0,$$ hence $$\dfrac{2}{\sqrt{(y^2+1)(z^2+1)}}\ge\dfrac{2}{\sqrt{(y+z)^2+1}}.$$ Therefore, it suffices to prove that $$\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}\ge1+\dfrac{1}{(y+z)^2+1}.$$ And this is equivalent to $$\dfrac{yz[2-2yz-yz(y+z)^2]}{(y^2+1)(z^2+1)[(y+z)^2+1]}\ge 0.$$ Above is true because $$2-2yz-yz(y+z)^2=2x(y+z)-yz(y+z)^2=(y+z)[2x-yz(y+z)]\ge (y+z)[2x-x^2(y+z)]=x(y+z)(2-xy-xz)\ge 0.$$

Lemma 2: $$\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}\ge\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}.$$

Proof: \begin{align}\dfrac{1}{x+y}+\dfrac{1}{\sqrt{x+z}}&=\sqrt{y+z}\left(\dfrac{1}{\sqrt{(y+z)(x+y)}}+\dfrac{1}{\sqrt{(z+x)(z+y)}}\right)\\ &=\sqrt{y+z}\left(\dfrac{1}{\sqrt{1+y^2}}+\dfrac{1}{\sqrt{1+z^2}}\right) \end{align} and apply Lemma 1, then done!

Thus the original inequality is equivalent to $$\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}+\sqrt{\dfrac{y+z}{(y+z)^2+1}}\ge 2+\dfrac{1}{\sqrt{2}}.$$

Letting $$t=\dfrac{1}{\sqrt{y+z}}+\sqrt{y+z}\ge 2,$$ the inequality becomes $$t+\dfrac{1}{\sqrt{t^2-2}}\ge 2+\dfrac{1}{\sqrt{2}}.$$

This can be shown by the following: \begin{align} &t+\dfrac{1}{\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}} \\ =&t+\dfrac{2\sqrt{2}}{2\sqrt{2}\cdot\sqrt{t^2-2}}-2-\dfrac{1}{\sqrt{2}} \\ \ge& t+\dfrac{2\sqrt{2}}{2+t^2-2}-2-\dfrac{1}{\sqrt{2}} \\ =&t+\dfrac{2\sqrt{2}}{t^2}-2-\dfrac{1}{\sqrt{2}} \\ =&\dfrac{(t-2)(\sqrt{2}t^2-t-2)}{\sqrt{2}t^2}\ge 0. \end{align}

  • $\begingroup$ I edited the format and some phrases. +1 $\endgroup$ – Shuhao Cao Jun 19 '13 at 14:47

$z=\dfrac{1-xy}{x+y} \ge 0 \to xy\le 1$


$f'_{x}=\dfrac{1-x^2-2xy}{(x^2+1) \sqrt{x^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(y^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<1>

$f'_{y}=\dfrac{1-y^2-2xy}{(y^2+1) \sqrt{y^2+1} \sqrt{x+y}}+\dfrac{1}{\sqrt{(x+y)(x^2+1)}}-\dfrac{1}{(x+y)\sqrt{x+y}}=0$......<2>



so we have $x=y$ only. put in <1>, we get: $4x-4x^3=(x^2+1)^{\frac{3}{2}}$,ie:

$15x^6-35x^4+13x^2-1=0 \to (3x^2-1)(5x^4-10x^2+1)=0$, we get two roots:

$x_1=\dfrac{1}{\sqrt{3}}, x_2=\sqrt{1-\dfrac{2}{\sqrt{5}}}$

$f(x_1)=2.79,f(x_2)=2.77$,but we note $f(x,y)$ when $x ,y \to \infty, f(x,y) \to 0$,so we have to check the bound, which is $xy=1$.


let $m=\sqrt{x}+\dfrac{1}{\sqrt{x}} \ge2$

$f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$, when $m\ge 2$, $f(m)$ is mono increasing, so the minimum will be got when $m=2 \to x=y=1, f(1)=2+\dfrac{1}{\sqrt{2}}< 2.77$, so the minimum is $2+\dfrac{1}{\sqrt{2}}$.

To prove $f(m)=m+\dfrac{1}{\sqrt{m^2-2}}$ is mono increasing function,

$f'(m)=1-\dfrac{m}{(m^2-2)^{3/2}}=1-\dfrac{1}{\sqrt{m^2-2}\left(m-\dfrac{2}{m}\right)}>0$ when $m\ge 2$.

  • $\begingroup$ Why are you able to restrict $z$ as $f(x, y)$ in the beginning? I thought $z$ was an independent variable $\endgroup$ – Ovi Jun 9 '13 at 19:43
  • $\begingroup$ I am not restrict z, it is just same function. you can use Lagrange multiplier to get same roots but it is very hard, especially for last root (1,1,0).it is mess. $\endgroup$ – chenbai Jun 10 '13 at 2:44

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