Prove $\color{black}{\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+b+2}}+\frac{1}{\sqrt{a+c+2}}\ge1+\frac{1}{\sqrt{2(a+b+c-1)}}.}$

Question

Question. Let $a,b,c\ge 0: a+b+c+abc=4.$ Prove that$$\color{black}{\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+b+2}}+\frac{1}{\sqrt{a+c+2}}\ge1+\frac{1}{\sqrt{2(a+b+c-1)}}.}$$

Because equality holds at $a=b=c=1,$ I tried to use Cauchy-Schwarz as $$\frac{1}{\sqrt{a+b+2}}+\frac{1}{\sqrt{c+b+2}}+\frac{1}{\sqrt{a+c+2}}\ge \frac{9}{\sqrt{a+b+2}+\sqrt{c+b+2}+\sqrt{a+c+2}}.$$

Also by C-S $$\sqrt{a+b+2}+\sqrt{c+b+2}+\sqrt{a+c+2}\le \sqrt{6(a+b+c+3)}.$$ Thus, the rest is just proving $$\frac{9}{\sqrt{6(a+b+c+3)}}\ge 1+\frac{1}{\sqrt{2(a+b+c-1)}},$$ which is wrong since $a+b+c\ge 3.$

I hope there are some better approachs. Thanks for contributors.

Answer 1

By Holder $$\left(\sum_{cyc}\frac{1}{\sqrt{a+b+2}}\right)^2\sum_{cyc}(a+b+2)((\sqrt6-1)c+a+b)^3$$ $$=\geq(\sqrt6+1)^3(a+b+c)^3=(19+9\sqrt6)(a+b+c)^3.$$ Thus, it's enough to prove that: $$(19+9\sqrt6)(a+b+c)^3\geq\left(1+\frac{1}{\sqrt{2(a+b+c-1)}}\right)^2\sum_{cyc}(a+b+2)((\sqrt6-1)c+a+b)^3,$$ which saves the case for the equality occurring: $a=b=c=1$ and $a=b=2$ with $c=0$.

I can prove the last inequality for $b=a$ and $c=\frac{4-2a}{a^2+1}.$

It's enough to check, what happens with $v^2=\frac{ab+ac+bc}{3}$ because the condition does not depend on $v^2$.

Can you end it now?

Answer 2

Proof.

We apply Ji Chen's Symmetric Function Theorem for $n=3$:
Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying $$x+y+z \ge u+v+w, \quad xy+yz+zx \ge uv+vw+wu, \quad xyz \ge uvw.$$ Then $x^d + y^d+z^d \ge u^d + v^d+w^d$. (See more information in here, here, and here.)

Let $$x = \frac{1}{a + b + 2}, \quad y = \frac{1}{b + c + 2}, \quad z = \frac{1}{c + a + 2}, $$ $$u = \frac14, \quad v = \frac14, \quad w = \frac{1}{2(a + b + c - 1)}.$$ The desired inequality is written as $$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge \sqrt{u} + \sqrt{v} + \sqrt{w}.$$

By Ji Chen's Symmetric Function Theorem, it suffices to prove that $$x + y + z \ge u + v + w, \tag{1}$$ and $$xy + yz + zx \ge uv + vw + wu, \tag{2}$$ and $$xyz \ge uvw. \tag{3}$$ The proofs of (1) and (2) are given at the end. (3) is true since $$(xy + yz + zx - uv - vw - wu) = (2a + 2b + 2c + 6)(xyz - uvw).$$

We are done.

$\phantom{2}$

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Proof of (1):

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a + b + c + abc = 4$ is written as $p + r = 4$. So $r = 4 - p$. Clearly, we have $p \le 4$. Using $p^3 \ge 27r$, we have $p^3/27 \ge 4 - p$ which results in $p \ge 3$. Thus, $3 \le p\le 4$.

We need to prove that $$-{p}^{2}q+6\,{p}^{2}+pr-2\,q-24 \ge 0. \tag{A1}$$

Using $r = 4-p$, (A1) is written as $$5p^2 + 4p - 24 - (p^2 + 2)q \ge 0. \tag{A2}$$

Using $r = 4- p$ and degree three Schur inequality $r \ge \frac{4pq - p^3}{9}$, we have $$q \le \frac{p^3 - 9p + 36}{4p}. \tag{A3}$$

From (A2) and (A3), it suffices to prove that $$5p^2 + 4p - 24 - (p^2 + 2)\cdot \frac{p^3 - 9p + 36}{4p} \ge 0$$ or $$\frac{(p - 3)(4 - p)(p^3 + 7p^2 + 10p + 6)}{4p} \ge 0$$ which is true. We are done.

Proof of (2):

Let $p = a + b + c, q = ab + bc + ca, r = abc$.

The condition $a + b + c + abc = 4$ is written as $p + r = 4$. So $r = 4 - p$. Clearly, we have $p \le 4$. Using $p^3 \ge 27r$, we have $p^3/27 \ge 4 - p$ which results in $p \ge 3$. Thus, $3 \le p\le 4$.

We need to prove that $$-2\,{p}^{2}-pq+24\,p-2\,q+r-40 \ge 0. \tag{B1}$$

Using $r = 4-p$, (B1) is written as $$-2p^2 + 23p - 36 - (p + 2)q \ge 0. \tag{B2}$$

Using $r = 4- p$ and degree three Schur inequality $r \ge \frac{4pq - p^3}{9}$, we have $$q \le \frac{p^3 - 9p + 36}{4p}. \tag{B3}$$

From (B2) and (B3), it suffices to prove that $$-2p^2 + 23p - 36 - (p + 2)\cdot \frac{p^3 - 9p + 36}{4p} \ge 0$$ or $$\frac{(p - 3)(4 - p)(p^2 + 17p + 6)}{4p} \ge 0$$ which is true. We are done.