Prove the inequality $xyz \geq xy+yz+xz \implies \sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}$

Let $x,y,x>0$ and $xyz \geq xy+yz+xz.$ Prove that $$\sqrt{x y z } \geq \sqrt{x}+\sqrt{y}+\sqrt{z}.$$

Solution.

Using AM GM inequality we have \begin{gather*} x y+xz \geq 2 \sqrt{x y x z}=2 x \sqrt{yz},\\ xy+yz \geq 2 y \sqrt{x z},\\ x z +y z \geq 2 z \sqrt{xz}. \end{gather*} Add and get $$xy+xz+yz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}.$$ By condition $$xyz \geq x \sqrt{yz}+y \sqrt{x z}+z \sqrt{xz}$$ Dividing by $\sqrt{xyz}$ we obtain the inequality.

Question. Are there another ways to prove it?

• I'm sure there are, but this method looks pretty good! – Dr Xorile Nov 16 '16 at 17:21
• Nice solution, I think there are other solutions but quite similar. – arberavdullahu Nov 16 '16 at 17:27
• It seem to me there is a slight misprint in your formulation of a problem. (maybe $\sqrt{xyz}$ ?) – kolobokish Nov 16 '16 at 17:42
• @ kolobokish, yes, fixed – Leox Nov 16 '16 at 18:22

From your assumption, we obtain that \begin{align*} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \le 1. \end{align*} Since \begin{align*} \frac{1}{x}+\frac{1}{y}+\frac{1}{z} &=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z} +\frac{1}{y} + \frac{1}{z} \right)\\ &\ge\frac{1}{2}\left(\frac{2}{\sqrt{xy}} + \frac{2}{\sqrt{xz}} +\frac{2}{\sqrt{yz}} \right), \end{align*} that is, \begin{align*} \frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{xz}} +\frac{1}{\sqrt{yz}}\le 1, \end{align*} the required inequality follows immediately by multiplying $\sqrt{xyz}$ to both sides.

Here's a proof that avoids (or disguises) the AM-GM inequality.

Let's start by establishing a completely different inequality: If $r,s\ge1$, then

$${1\over s}+r^2s+s\ge1+r+rs$$

The proof comes from first rewriting it in equivalent form as

$$(r-1)rs^2+(s-1)s\ge rs-1$$

and then letting $r=1+a$ and $s=1+b$ with $a,b\ge0$:

$$a(1+a)(1+b)^2+b(1+b)\ge a+b+ab$$

where we've expanded the right hand side but not the left. However, since there are no negative signs on the left, it's easy to see that its expansion consists exclusively of non-negative terms, including an $a$, a $b$, and an $ab$ (in fact a $2ab$), hence cannot be less than $a+b+ab$. This completes the proof.

Now to the problem about $x,y,z$. The statement is symmetric in the three variables, so we can, without loss of generality, assume $0\lt x\le y\le z$. This allows us to write $x=u^2$, $y=r^2u^2$ and $z=r^2s^2u^2$ with $u\gt0$ and $r,s\ge1$. Then

\begin{align} xyz\ge xy+yz+zx&\implies r^4s^2u^6\ge r^2u^4+r^4s^2u^4+r^2s^2u^4\\ &\implies r^2su^3\ge{u\over s}+r^2su+su= \left({1\over s}+r^2s+s\right)u\\ &\implies r^2su^3\ge(1+r+rs)u=u+ru+rsu\\ &\implies\sqrt{xyz}\ge\sqrt x+\sqrt y+\sqrt z \end{align}

My another solution inspired by Gordon's solution.

Suppose that $x \leq y \leq z$ Then $\sqrt{x} \leq \sqrt{y} \leq \sqrt{z}$ and
$$\frac{1}{\sqrt{x}} \geq \frac{1}{\sqrt{y}} \geq \frac{1}{\sqrt{z}}.$$ Now by rearrangement inequality we get $$1 \geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \geq \frac{1}{\sqrt{xy}} + \frac{1}{\sqrt{xz}} +\frac{1}{\sqrt{yz}}.$$

• That is usual rearrangenent inequality – Leox Nov 17 '16 at 18:53

You could use this famous inequality: $$xy+yz+xz\geq \sqrt{3xyz(x+y+z)}$$ (which can be easily proved by AM-GM).

Using that inequality we have $$xyz\geq xy+yz+zx\geq \sqrt{3xyz(x+y+z)},$$ from where is $$\sqrt{xyz}\geq \sqrt{3(x+y+z)}=\sqrt{(1+1+1)(x+y+z)}$$ By Cauchy-Schwartz inequality, we have $$\sqrt{(1+1+1)(x+y+z)} \geq (\sqrt{x}+\sqrt{y}+\sqrt{z})$$ and finally we can conclude $$\sqrt{xyz}\geq\sqrt{x}+\sqrt{y}+\sqrt{z}.$$