# Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$

Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$

• why dont you try removing the square root from the denominator and then use Cauchy Schwartz? Dec 30, 2014 at 13:52

## 3 Answers

By AM-HM or C-S we have $$\frac{\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}}{3} \geq \frac{3}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$

Therefore $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}} \geq \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx}}$$

To prove your claim, you need to show $$\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \leq \sqrt{10}$$

By C-S $$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \right)^2 \leq 3 (1+xy+1+yz+1+zx )=9+3(xy+xz+yz)$$

Finally, again by C-S we have $$xy+xz+yz \leq x^2+y^2+z^2 \Rightarrow 3\\(xy+xz+yz) \leq x^2+y^2+z^2+2(xy+xz+yz) =(x+y+z)^2=1$$

Combining the last two inequalities we get $$\left(\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+zx} \right)^2 \leq 10$$

By the inequality between harmonic and arithmetic means we have that $$\frac{1}{\sqrt{1+xy}}+\frac{1}{\sqrt{1+yz}}+\frac{1}{\sqrt{1+xz}}\ge \frac{9}{\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}}.$$ Now if we prove that $$\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{10}$$ we are done. but by Cauchy-Schwarz we have $$\sqrt{1+xy}+\sqrt{1+yz}+\sqrt{1+xz}\le\sqrt{3}(3+xy+yz+zx)^{\frac{1}{2}},$$ so we just need to prove that $$xy+yz+zx\le\frac{1}{3},$$ but it is the same to prove that $$x^2+y^2+z^2\ge\frac{1}{3}.$$ The last inequality can be easily proved using Cauchy-Schwarz.

• There is no final e in "Cauchy-Schwarz" Dec 30, 2014 at 14:23

Here is another approach.

Cauchy-Schwarz says \begin{align} (x+y+z)^2 &\le(x^2+y^2+z^2)(1^2+1^2+1^2)\\ &=3(x^2+y^2+z^2)\tag{1} \end{align} Therefore, \begin{align} xy+yz+zx &=\tfrac12\left[(x+y+z)^2-\left(x^2+y^2+z^2\right)\right]\\ &\le\tfrac12\left[(x+y+z)^2-\tfrac13(x+y+z)^2\right]\\ &=\tfrac13(x+y+z)^2\tag{2} \end{align} Then \begin{align} \frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}} &\ge\frac3{\sqrt[\large6]{(1+xy)(1+yz)(1+zx)}}\tag{3}\\[3pt] &\ge\frac3{\sqrt{\frac13\left[(1+xy)+(1+yz)+(1+zx)\right]}}\tag{4}\\ &=\frac9{\sqrt{9+3(xy+yz+zx)}}\tag{5}\\[3pt] &\ge\frac9{\sqrt{10}}\tag{6} \end{align} Explanation:
$(3)$: AM-GM
$(4)$: AM-GM in the denominator inside the square root
$(5)$: regrouping
$(6)$: apply $(2)$