Prove $\sum_{cyc}xy\left(2\sqrt{z^2+1}-z\right)\ge\sum_{cyc}\left(\sqrt{z^2+1}-z\right)$ when $xy+yz+zx=1$

Question

Given $x,y,z$ be non negative real numbers satisfying $xy+yz+zx=1.$ Prove that $$\sum_{cyc}xy\left(2\sqrt{z^2+1}-z\right)\ge\sum_{cyc}\left(\sqrt{z^2+1}-z\right). $$ My thoughts is proving$$xy\left(2\sqrt{z^2+1}-z\right)\ge\sqrt{z^2+1}-z. $$ Or $$xy\ge\frac{\sqrt{z^2+1}-z}{2\sqrt{z^2+1}-z}.$$It is clearly true because $xy\ge 0\ge\dfrac{\sqrt{z^2+1}-z}{2\sqrt{z^2+1}-z}$ and we end proof.

The inequality $$0\ge\dfrac{\sqrt{z^2+1}-z}{2\sqrt{z^2+1}-z}$$ is true since $2\sqrt{z^2+1}>\sqrt{z^2+1}$ and $2\sqrt{z^2+1}>0.$

Is there other solution? I am appreciate your contributions.

Answer 1

Proof.

By replacing $xy+yz+zx=1,$ the OP is equivalent to $$\sum_{cyc}\sqrt{(x+y)(x+z)}(xy+xz-yz)\le \sum_{cyc}yz(y+z),$$ $$\iff \sum_{cyc}yz\sqrt{(z+x)(x+y)}+(x+y)(y+z)(z+x)\ge \sum_{cyc}(xy+xz)\sqrt{(z+x)(x+y)}+2xyz.$$ Dividing both side by $(x+y)(y+z)(z+x),$ we obtain $$\sum_{cyc}\dfrac{xy}{(x+y)\sqrt{(z+y)(x+z)}}+1\ge \sum_{cyc}\dfrac{x}{\sqrt{(x+y)(x+z)}}+\frac{2xyz}{(x+y)(y+z)(z+x)}.$$ Now, let $a=\dfrac{x}{\sqrt{(x+y)(x+z)}};b=\dfrac{y}{\sqrt{(x+y)(y+z)}};c=\dfrac{z}{\sqrt{(z+y)(x+z)}},$ we will prove

Problem. If $a,b,c\ge 0:a^2+b^2+c^2+2abc=1.$ Prove that$$ab+bc+ca+1\ge a+b+c+2abc.$$

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Notice that:$$ab+bc+ca+1\ge a+b+c+2abc\iff (1-a)(1-b)(1-c)\ge abc. \tag{*}$$ The remain is proving $(*)$ true.

Indeed, by the hypothesis$$(a+bc)^2=(1-b^2)(1-c^2),$$ $$(b+ca)^2=(1-c^2)(1-a^2),$$ $$(a+bc)^2=(1-b^2)(1-c^2).$$ Multiply three above inequalities, we obtain$$(a+bc)(b+ca)(c+ab)=(1-a^2)(1-b^2)(1-c^2). \tag{1}$$ Also, $(a+bc)(b+ca)-ab(c+1)^2=c(a-b)^2\ge 0,$ hence $$(a+bc)(b+ca)\ge ab(c+1)^2,$$ $$(b+ac)(c+ba)\ge bc(a+1)^2,$$ $$(c+ba)(a+cb)\ge ac(b+1)^2.$$It implies $$(a+bc)(b+ca)(c+ab)\ge abc(a+1)(b+1)(c+1).\tag{2}$$From $(1)$ and $(2)$, we proved the following inequality$$(1-a^2)(1-b^2)(1-c^2)\ge abc(a+1)(b+1)(c+1),$$or $(*)$ is true and the OP is proven. The proof ends here. Equality holds at $a=b=c=\dfrac{1}{2},$ or $x=y=z=\dfrac{\sqrt{3}}{3}.$

Answer 2

We need to prove that: $$\sum_{cyc}xy\left(2\sqrt{(x+z)(y+z)}-z\right)\geq(xy+xz+yz)\sum_{cyc}\left(\sqrt{(x+z)(y+z)}-z\right)$$ or $$(x+y+z)(xy+xz+yz)-3xyz\geq(xz+yz-xy)\sum_{cyc}\sqrt{(x+z)(y+z)}$$ or $$\sum_{cyc}(2x^2y+2x^2z-(xz+yz-xy)(x+z+y+z))+\sum_{cyc}(xz+yz-xy)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2\geq0$$ or $$\sum_{cyc}(xz+yz-xy)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2\geq0.$$ Now, let $x\geq y\geq z$.

Thus, $xy+yz-xz\geq0$ and $xy+xz-yz\geq0$ and it's enough to prove that: $$(xz+yz-xy)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2+(xy+yz-xz)\left(\sqrt{x+y}-\sqrt{y+z}\right)^2\geq0.$$ But $$\left(\sqrt{x+y}-\sqrt{y+z}\right)^2\geq\left(\sqrt{x+z}-\sqrt{y+z}\right)^2.$$ Thus, $$(xz+yz-xy)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2+(xy+yz-xz)\left(\sqrt{x+y}-\sqrt{y+z}\right)^2\geq$$ $$\geq(xz+yz-xy)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2+(xy+yz-xz)\left(\sqrt{x+z}-\sqrt{y+z}\right)^2=$$ $$=\left(\sqrt{x+z}-\sqrt{y+z}\right)^2(xz+yz-xy+xy+yz-xz)=2yz\left(\sqrt{x+z}-\sqrt{y+z}\right)^2\geq0.$$