If $xy+yz+zx+xyz=4,$ prove $\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{y+1}}+\frac{1}{\sqrt{z+1}} \leq 1+\frac{2}{\sqrt{3}}.$

Question

Problem. Let $x,y,z\ge 0: xy+yz+zx+xyz=4.$ Prove that$$\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{y+1}}+\frac{1}{\sqrt{z+1}} \leq 1+\frac{2}{\sqrt{3}}.$$

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I tried to use the substitution $x=\dfrac{2a}{b+c};y=\dfrac{2b}{a+c};z=\dfrac{2c}{b+a}.$ The original inequality becomes $$ \sqrt{\frac{a+b}{a+b+2 c}}+\sqrt{\frac{b+c}{b+c+2 a}}+\sqrt{\frac{c+a}{c+a+2 b}} \leq 1+\frac{2}{\sqrt{3}}, $$ for all $a,b,c\ge 0: a+b+c>0.$

Hope we can find some ideas. Thank you.

Answer 1

Remarks: Here is a proof motivated by computer. It takes some minutes by hand to verify the identity (1) (e.g. full expanding (1)).

We have the following identity: \begin{align*} f(a, b, c)\left(1 + \frac{2}{\sqrt 3} - \frac{1}{a + 1} - \frac{1}{b + 1} - \frac{1}{c + 1}\right) - g(a, b, c) = h(a, b, c) \tag{1} \end{align*} where $$f(a, b, c) := (6 + 4\sqrt 3)(a + 1)(b + 1)(c + 1)(ab + bc + ca + abc),$$ and \begin{align*} g(a, b, c) &:= (a^2 + 2a)(b^2 + 2b) + (b^2 + 2b)(c^2 + 2c) + (c^2 + 2c)(a^2 + 2a)\\ &\qquad + (a^2 + 2a)(b^2 + 2b)(c^2 + 2c) - 4, \end{align*} and \begin{align*} h(a, b, c) &:= \left( 14-4\,\sqrt {3} \right) abc+4( 2\,\sqrt {3} - 3 ) {a}^{2}{b}^{2}{c}^{2}+2\sqrt 3\, \left( ab+ac+bc \right) ab c\\ &\qquad + \Big( ( 2+\sqrt {3}) \left( ab+ac+bc \right) +5\,abc -2 \Big) ^{2}. \end{align*}

Letting $a = \sqrt{1 + x} - 1, b = \sqrt{1 + y} - 1, c = \sqrt{1 + z} - 1$, we have $a, b, c \ge 0$, and $f(a, b, c) > 0$, and $g(a, b, c) = xy + yz + zx + xyz - 4 = 0$, and $h(a, b, c) \ge 0$. From (1), we have $$1 + \frac{2}{\sqrt 3} - \frac{1}{\sqrt{x + 1}} - \frac{1}{\sqrt{y + 1}} - \frac{1}{\sqrt{z + 1}} \ge 0.$$

We are done.

Answer 2

For $z=0$ we obtain $xy=4$ and we need to prove that: $$\frac{1}{\sqrt{x+1}}+\frac{1}{\sqrt{y+1}}\leq\frac{2}{\sqrt3}$$ or $$\frac{1}{x+1}+\frac{1}{y+1}+\frac{2}{\sqrt{(x+1)(y+1)}}\leq\frac{4}{3}$$ or $$\frac{2}{\sqrt{x+y+5}}\leq\frac{x+y+14}{3(x+y+5)}$$ or $$x+y+14\geq6\sqrt{x+y+5},$$ which is true by AM-GM: $$x+y+14=x+y+5+9\geq2\sqrt{(x+y+5)\cdot9}=6\sqrt{x+y+5}.$$ Now, let $$f(x,y,z,\lambda)=\sum_{cyc}\frac{1}{\sqrt{x+1}}+\lambda(xy+xz+yz+xyz-4).$$ Thus, in the inside maximum point $(x,y,z)$ we need $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial y}=0$$ or $$-\frac{1}{2\sqrt{(x+1)^3}}+\lambda(y+z+yz)=-\frac{1}{2\sqrt{(y+1)^3}}+\lambda(x+z+xz)=-\frac{1}{2\sqrt{(z+1)^3}}+\lambda(x+y+xy)=0,$$ which gives $$(x+1)^3(y+z+yz)^2=(y+1)^3(x+z+xz)^2=(z+1)^3(x+y+xy)^2.$$ Now, let in the maximum point $x\neq y$ and $x\neq z$.

Thus, from $$(x+1)^3(y+z+yz)^2=(y+1)^3(x+z+xz)^2$$ we obtain $$(x-y)\left(x^2y^2z^2+2z^2xy+2z^2x+2z^2y-xy+z^2-z+\sum_{cyc}(2x^2y^2z+2z^2xy+x^2y^2-2xy-x)\right)=0$$ and from $$(x+1)^3(y+z+yz)^2=(z+1)^3(x+y+xy)^2$$ we obtain: $$(x-z)\left(x^2y^2z^2+2y^2xz+2y^2x+2y^2z-xz+y^2-y+\sum_{cyc}(2x^2y^2z+2z^2xy+x^2y^2-2xy-x)\right)=0,$$ which gives $$2y^2xz+2y^2x+2y^2z-xz+y^2-y=2z^2xy+2z^2x+2z^2y-xy+z^2-z$$ or $$(y-z)(2xyz+2xy+2xz+2yz+x+y+z-1)=0$$ or $$(y-z)(x+y+z+7)=0,$$ which gives $y=z$.

Id est, it's enough to prove our inequality for equality case of two variables.

Let $y=x$.

Thus, the condition gives $$x^2+2xz+x^2z=4$$ or $$xz(2+x)=4-x^2$$ or $$z=\frac{2-x}{x},$$ where $0<x\leq2.$

Thus, we need to prove that: $$\frac{2}{\sqrt{x+1}}+\frac{1}{\sqrt{1+\frac{2-x}{x}}}\leq1+\frac{2}{\sqrt3}$$ or $$\frac{2}{\sqrt{x+1}}-\frac{2}{\sqrt3}\leq1-\sqrt{\frac{x}{2}}$$ or $$\frac{2\sqrt2(2-x)}{\sqrt3\sqrt{1+x}(\sqrt3+\sqrt{1+x})}\leq\frac{2-x}{\sqrt2+\sqrt{x}}$$ or $$2\sqrt2(\sqrt2+\sqrt{x})\leq\sqrt3\left(\sqrt{3(1+x)}+1+x\right),$$ which is strong enough, but smooth.

Can you end it now?