How to prove $\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge \frac{\sqrt{abc+4}+4\sqrt{ab+bc+ca+4}}{2}.$

Question

Question. If $a,b,c\ge 0: a+b+c=2,$ prove that $$\color{blue}{\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge \frac{\sqrt{abc+4}+4\sqrt{ab+bc+ca+4}}{2}.}$$Equality holds iff $(a,b,c)=\{(0,1,1);(0,0,2)\}.$

Source: From a Israel Mathematical book.

My strategy is using Jichen lemma, but the first condition is already wrong.

Indeed, we can rewrite the original inequality as $$\color{black}{\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}\ge \sqrt{\frac{abc+4}{4}}+2\cdot\sqrt{ab+bc+ca+4}.}$$ Now, note that $$4(a+b+c)+3\ge \frac{abc+4}{4}+2(ab+bc+ca+4) $$ $$\iff 2\ge \frac{abc}{4}+2(ab+bc+ca),$$which is already wrong when $a=b=c=\dfrac{2}{3}.$

I think that Jichen lemma is a good approach to kill this kind of inequality but the lemma doesn't work actually.

I hope someone can find better ideas. All idea and comment are welcome.

Answer 1

For $abc=0$ the equality is true.

Let $abc\neq 0$ and $f(a,b,c)=\sum\limits_{cyc}\sqrt{4a+1}-\frac{\sqrt{abc+4}+4\sqrt{ab+bc+ca+4}}{2}+\lambda(a+b+c-2)$.

Let $(a,b,c)$ is an inside minimum point.

Thus, $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0,$$ which gives: $$\frac{2}{\sqrt{4a+1}}-\frac{bc}{4\sqrt{abc+4}}-\frac{b+c}{\sqrt{ab+ac+bc+4}}=$$ $$=\frac{2}{\sqrt{4b+1}}-\frac{ac}{4\sqrt{abc+4}}-\frac{a+c}{\sqrt{ab+ac+bc+4}}=$$ $$=\frac{2}{\sqrt{4c+1}}-\frac{ab}{4\sqrt{abc+4}}-\frac{a+b}{\sqrt{ab+ac+bc+4}}.$$ Now, let $a\neq b$ and $a\neq c$.

Thus, $$(a-b)\left(-\frac{8}{\sqrt{4a+1}+\sqrt{4b+1}}+\frac{c}{4\sqrt{abc+4}}+\frac{1}{\sqrt{ab+ac+bc+4}}\right)=0$$ and

$$(a-c)\left(-\frac{8}{\sqrt{4a+1}+\sqrt{4c+1}}+\frac{b}{4\sqrt{abc+4}}+\frac{1}{\sqrt{ab+ac+bc+4}}\right)=0,$$ which gives $$-\frac{8}{\sqrt{4a+1}+\sqrt{4b+1}}+\frac{c}{4\sqrt{abc+4}}=-\frac{8}{\sqrt{4a+1}+\sqrt{4c+1}}+\frac{b}{4\sqrt{abc+4}}$$ or $$(b-c)\left(\frac{32}{\prod\limits_{cyc}\sqrt{4a+1}}-\frac{1}{4\sqrt{abc+4}}\right)=0$$ and since by AM-GM and Jensen $$\frac{32}{\prod\limits_{cyc}\sqrt{4a+1}}\geq\frac{32}{\left(\frac{\sum\limits_{cyc}\sqrt{4a+1}}{3}\right)^3}\geq\frac{32}{\left(\sqrt{\frac{4(a+b+c)}{3}+1}\right)^3}>\frac{1}{8}>\frac{1}{4\sqrt{abc+4}},$$ we obtain $b=c$ and it's enough to prove our inequality for equality case of two variables.

Can you end it now?

Answer 2

Alternative proof: the pqr method

Let $x := \sqrt{4a + 1} - 1, y := \sqrt{4b + 1} - 1, z := \sqrt{4c + 1} - 1$. We have $a = \frac14x^2 + \frac12x, b = \frac14y^2 + \frac12y, c = \frac14z^2 + \frac12z$. Let $p = x + y + z, q = xy + yz + zx, r = xyz$.

The condition $a + b + c = 2$ is written as $\frac14p^2 + \frac12p - \frac12 q = 2$ which gives $$q = \frac12p^2 + p - 4.\tag{1}$$

From (1) and $q \ge 0$, we have $p \ge 2$. From (1) and $p^2 \ge 3q$, we have $p \le \sqrt{33} - 3 $. Thus, $$2 \le p \le \sqrt{33} - 3. \tag{2}$$

Using $p^3 \ge 27r$ and (2), we have $r \le p^3/27 < 1$. From degree three Schur inequality, we have $r \ge \frac{4pq - p^3}{9} = \frac{p^3 + 4p^2 - 16p}{9}$. Thus, we have $$\max(0, r_1) \le r < 1 \tag{3}$$ where $r_1 := \frac{p^3 + 4p^2 - 16p}{9}$.

We need to prove that, under the conditions (2) and (3), $$p + 3 \ge \frac{\sqrt{p^2r + 6pr + r^2 + 256}}{16} + \frac{\sqrt{p^4+8p^3+4p^2-8pr-48p-24r+256}}{4}.\tag{4}$$

Let $f(r) := \mathrm{LHS}_{(4)} - \mathrm{RHS}_{(4)}$. Using (2), we have, for all $0 \le r < 1$, \begin{align*} f'(r) &= \frac{p + 3}{\sqrt{p^4+8p^3+4p^2-8pr-48p-24r+256}} - \frac{p^2 + 6p + 2r}{32\sqrt{p^2r + 6pr + r^2 + 256}}\\[6pt] &\ge \frac{p + 3}{\sqrt{p^4+8p^3+4p^2+256}} - \frac{p^2 + 6p + 2\cdot 1}{32\sqrt{256}}\\[6pt] &> 0. \end{align*}

We split into two cases.

Case 1. $2 \le p \le 2(\sqrt 5 - 1)$

We have $r_1 \le 0$. (3) becomes $0 \le r < 1$.

We have $$f(0) = p + 2 - \frac14\sqrt{p^4 + 8p^3 + 4p^2 - 48p + 256} \ge 0.$$ Thus, $f(r) \ge 0$ for all $0 \le r < 1$.

Case 2. $2(\sqrt 5 - 1) < p \le \sqrt{33} - 3$

We have $r_1 > 0$. (3) becomes $r_1 \le r < 1$.

We have \begin{align*} f(r_1) &= p + 3 - \frac{1}{144} \sqrt{p^6+17p^5+74p^4-56p^3-608p^2+20736}\\[6pt] &\qquad - \frac{1}{12}\sqrt{p^4+16p^3+68p^2-48p+2304}\\[6pt] &\ge 0. \end{align*} Thus, $f(r) \ge 0$ for all $r_1 \le r < 1$.

We are done.