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.
Best Answer
Without loss of generality, we assume that $x\ge y\ge z$.
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 $$2-2yz-yz(y+z)^2=2x(y+z)-yz(y+z)^2=(y+z)[2x-yz(y+z)]\ge (y+z)[2x-x^2(y+z)]=x(y+z)(2-xy-xz)\ge 0.$$
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}