Prove $\color{black}{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\cdot\sqrt{\frac{a+b+c+5}{3}}, }$ when $ab+bc+ca+abc=4.$

Question

If $a,b,c\ge 0: ab+bc+ca+abc=4.$ Prove that$$\color{black}{\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}\le 1+2\cdot\sqrt{\frac{a+b+c+5}{3}}. }$$ I've tried to square both side but it leads to complicated one.

Now, if we use the substitution $a=\dfrac{2x}{y+z};b=\dfrac{2y}{x+z};c=\dfrac{2z}{y+x}.$ The problem turns out $$\sum_{cyc}\sqrt{\frac{2x+y+z}{y+z}}\le 1+\frac{2}{\sqrt{3}}\cdot\sqrt{5+2\sum_{cyc}\frac{x}{y+z}}.$$ From here, I don't know how to continue to full proof. Hope you can share some thoughts to help me out.

Also, all idea and comment are welcome. Thanks for interest.

Answer 1

Let $c=0$.

Thus, $ab=4$ and the inequality it's $$2\sqrt{\frac{a+b+5}{3}}\geq\sqrt{a+1}+\sqrt{b+1}$$ or $$\frac{4(a+b+5)}{3}\geq a+b+2+2\sqrt{a+b+5}$$ $$a+b+14\geq6\sqrt{a+b+5}$$ or

$$\left(\sqrt{a+b+5}-3\right)^2\geq0.$$

Now, let $$f(a,b,c)=1+2\sqrt{\frac{a+b+c+5}{3}}-\sum_{cyc}\sqrt{a+1}+\lambda(ab+ac+bc+abc-4).$$

Thus, in the inside minimum point we have: $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives $$-\frac{1}{2\sqrt{a+1}}+\lambda(b+c+bc)=-\frac{1}{2\sqrt{b+1}}+\lambda(a+c+ac)=-\frac{1}{2\sqrt{c+1}}+\lambda(a+b+ab).$$ From the first equality we obtain: $$\frac{1}{\sqrt{b+1}}-\frac{1}{\sqrt{a+1}}=2\lambda(a+ac-b-bc)$$ or $$(a-b)\left(\frac{1}{\sqrt{(a+1)(b+1)}(\sqrt{a+1}+\sqrt{b+1})}-2\lambda(c+1)\right)=0.$$ By the similar way we obtain $$(a-c)\left(\frac{1}{\sqrt{(a+1)(c+1)}(\sqrt{a+1}+\sqrt{c+1})}-2\lambda(b+1)\right)=0.$$
Now, let in this point be $a\neq b$ and $a\neq c$.

Thus, $$\sqrt{(a+1)(b+1)}(\sqrt{a+1}+\sqrt{b+1})(c+1)=\sqrt{(a+1)(c+1)}(\sqrt{a+1}+\sqrt{c+1})(b+1)$$ or $$\sqrt{b+1}(\sqrt{a+1}+\sqrt{b+1})(c+1)=\sqrt{c+1}(\sqrt{a+1}+\sqrt{c+1})(b+1)$$ $$\sqrt{b+1}=\sqrt{c+1},$$ which gives $b=c$.

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

Let $b=a$.

Thus, the condition gives $c=\frac{2-a}{a},$ where $0<a<2$ and we need to prove that: $$2\sqrt{a+1}+\sqrt{\frac{2}{a}}\leq1+\frac{2\sqrt2(a+1)}{\sqrt{3a}}.$$ Now, let $a=x^2,$ where $0<x<\sqrt2.$

Thus, we need to prove that: $$2\sqrt2x^2+\sqrt3x+2\sqrt2-\sqrt6\geq2x\sqrt{3(x^2+1)}$$ or $$(\sqrt2-x)\left(7\sqrt2-4\sqrt6+x-4(\sqrt6-\sqrt2)x^2+4x^3\right)\geq0,$$ which is true for $0<x<\sqrt2$.

Answer 2

Proof.

Let $x := \sqrt{a + 1} - 1, y := \sqrt{b + 1} - 1, z := \sqrt{c + 1} - 1$.
We have $a = x^2 + 2x, b = y^2 + 2y, c = z^2 + 2z$.

Let $p = x + y + z, q = xy + yz + zx, r = xyz$.

The condition $ab + bc + ca + abc = 4$ is written as $$r^2 + (2p + 2q + 2)r + 2pq + q^2 + 4q - 4 = 0. \tag{1}$$

We need to prove that $$p + 3 \le 1 + 2\sqrt{\frac{p^2 + 2p - 2q + 5}{3}}. \tag{2}$$

From (1), we have $2pq + q^2 + 4q - 4 \le 0$ which results in $$q \le \sqrt{p^2 + 4p + 8} - p - 2. \tag{3}$$

From (2) and (3), it suffices to prove that $$p + 3 \le 1 + 2\sqrt{\frac{p^2 + 2p - 2\cdot (\sqrt{p^2 + 4p + 8} - p - 2) + 5}{3}}$$ or $$(p + 2)^2 \le 4\cdot \frac{p^2 + 2p - 2\cdot (\sqrt{p^2 + 4p + 8} - p - 2) + 5}{3}$$ or $$(p^2 + 4p + 8) + 16 \ge 8\sqrt{p^2 + 4p + 8}$$ which is true by AM-GM.

We are done.