Finding $\small{\max\limits_{ab+bc+ca=1}\sqrt{\frac{1}{a+1}+\frac{1}{b+1}}+\sqrt{\frac{1}{a+1}+\frac{1}{c+1}}+\sqrt{\frac{1}{c+1}+\frac{1}{b+1}}.}$

Question

Let $a,b,c\ge 0: ab+bc+ca=1.$ Find the maximum $$P=\sqrt{\frac{1}{a+1}+\frac{1}{b+1}}+\sqrt{\frac{1}{a+1}+\frac{1}{c+1}}+\sqrt{\frac{1}{c+1}+\frac{1}{b+1}}.$$ By denote some specific value, I think the maximal is $1+\sqrt{6}$ achieved at $(a,b,c)=(0,1,1).$

Now, by using Cauchy-Schwarz inequality, $$P\le \sqrt{3}.\sqrt{2\left(\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\right)}.$$But it seems not good enough. I was in trouble to apply classical approach according to occuring equality.

Hope to see some ideas. I am very appreciate your help.

Answer 1

Some thoughts.

We can use Jichen lemma.

Indeed, we can rewrite the original inequality as$$\sqrt{\frac{1}{a+1}+\frac{1}{b+1}}+\sqrt{\frac{1}{a+1}+\frac{1}{c+1}}+\sqrt{\frac{1}{c+1}+\frac{1}{b+1}}\le 1+\sqrt{\frac{3}{2}}+\sqrt{\frac{3}{2}}.$$ By the lemma, we just need to prove the following inequalities $$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\le 2 \tag{1}$$ $$\sum_{cyc}{\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\left(\frac{1}{c+1}+\frac{1}{b+1}\right)}\le \frac{21}{4} \tag{2}$$$$\left(\frac{1}{a+1}+\frac{1}{b+1}\right)\left(\frac{1}{c+1}+\frac{1}{b+1}\right)\left(\frac{1}{c+1}+\frac{1}{a+1}\right)\le \frac{9}{4} \tag{3}$$ The rest is smooth. Can you end it now?

Answer 2

According to Mathematica, the determinant of the Hessian matrix of $(x,y) \mapsto \sqrt{\frac{1}{1+x}+\frac{1}{1+y}}$ is $$ \frac{3}{4 (x+1)^2 (y+1)^2 (x+y+2)}$$

Its trace is $$\frac{3 x^4+4 x^3 (y+4)+6 x^2 (2 y+5)+4 x \left(y^3+3 y^2+6 y+7\right)+3 y^4+16 y^3+30 y^2+28 y+14}{4 (x+1)^3 (y+1)^3 (x+y+2) \sqrt{\frac{x+y+2}{x y+x+y+1}}}$$

They are positive for $x,y \in [0,1]$, showing that the function is convex. Therefore $$(a,b,c) \mapsto \sum \sqrt{\frac{1}{1+a}+\frac{1}{1+b}}$$ is also convex.

The key now is understanding that moving variables apart increases the objective function. Obviously $a,b,c \in [0,1]$. Suppose that $0<a,b,c<1$ and $a\leq b$. Replacing $a,b$ with $a-\epsilon, b+\epsilon$ gives $$ (a-\epsilon)(b+\epsilon)+(b+\epsilon)c+c(a-\epsilon) = 1-\epsilon^2+\epsilon(b-a).$$

For $\epsilon>0$ small enough the value of the constraint function is increased so to preserve equality we must choose $c_\epsilon<c$ such that $$ (a-\epsilon)(b+\epsilon)+(b+\epsilon)c_\epsilon+c_\epsilon(a-\epsilon) = 1$$

Convexity shows that moving $a,b$ apart increases the objective function. Decreasing the variable $c$ further increases the objective function. Thus if all $a,b,c$ are strictly between $0$ and $1$ they do not maximize the function.

Thus, at least one of $a,b,c$ belongs to $\{0,1\}$. Continuing the casework will lead to the solution.

This is not an elegant solution, but hopefully it will lead to one.

Answer 3

Alternative proof.

WLOG, assume that $a \ge b \ge c$. We split into two cases.

Case 1. $a < 3$

Let $A := \frac{1}{a+1} + \frac{1}{b+1}, B := \frac{1}{b+1} + \frac{1}{c+1}, C := \frac{1}{c+1} + \frac{1}{a+1}$. Using $\sqrt{u} + \sqrt{v} \le \sqrt{2(u+v)}$, it suffices to prove that $$\sqrt{A} + \sqrt{2(B + C)} \le 1 + \sqrt{6}$$ or $$\frac{A - 1}{\sqrt{A} + 1} \le \frac{6 - 2(B + C)}{\sqrt{6} + \sqrt{2(B+C)}}$$ or $$\frac{A - 1}{1 + 1} \le \frac{6 - 2(B + C)}{5}$$ or $$\frac{17abc + 9ab + 8bc + 8ca - c - 9}{10(a+1)(b+1)(c+1)}\ge 0 \tag{1}$$ where we use $A - 1 = \frac{1 - ab}{(a+1)(b+1)}\ge 0$, and $6 - 2(B+C) = \frac{2(3abc + ab + 2bc + 2ca + c - 1)}{(a+1)(b+1)(c+1)} \ge 0$, and $2\sqrt{6} < 5$.

(1) is true. Indeed, using $a \ge \frac{1}{\sqrt 3}$ and $ab \ge \frac13$ (easy), we have \begin{align*} 17abc + 9ab + 8bc + 8ca - c - 9 &= 17abc- bc - ca - c\\ &= c(17ab - a - b - 1)\\ &= c[(17a - 1)b - a - 1]\\ &\ge c\left[(17a - 1)\cdot \frac{1}{3a} - a - 1\right]\\ &\ge 0. \end{align*}

Case 2. $a \ge 3$

Using $\sqrt{u} + \sqrt{v} \le \sqrt{2(u+v)}$, we have \begin{align*} P &\le \sqrt{2\left(\frac{1}{a+1}+\frac{1}{b+1} + \frac{1}{a+1}+\frac{1}{c+1}\right)} + \sqrt{\frac{1}{b+1}+\frac{1}{c+1}}\\[6pt] &\le \sqrt{2\left(\frac{2}{a+1}+ \frac{1 + 2a}{1 + a}\right)} + \sqrt{\frac{1 + 2a}{1 + a}}\\[6pt] &= \sqrt{4 + \frac{2}{1 + a}} + \sqrt{2 - \frac{1}{1 + a}}\\[6pt] &< 1 + \sqrt 6 \end{align*}
where we use $\frac{1 + 2a}{1 + a} - \frac{1}{b+1}-\frac{1}{c+1} = \frac{2abc + ab + bc + ca - 1}{(1 + a)(1 + b)(1 + c)} = \frac{2abc}{(1 + a)(1 + b)(1 + c)} \ge 0$, and $\sqrt{4 + 2x} + \sqrt{2 - x} < 1 + \sqrt 6$ for all $0 \le x \le \frac{1}{4}$ (simply letting $x = \frac{1}{1 + a}$).

We are done.