If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$.

Question

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$

The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.

I tried AM-GM, C-S and more, but without any success.

P.S.

I want to show here a refinement of the beautiful River li's solution.

We need to prove that: $$\sum_{\mathrm{cyc}}\frac{2a+1}{4a + 2b + 3}\sum_{\mathrm{cyc}}\frac{a}{2a+1}\le 1$$ or $$\frac{\sum\limits_{cyc}(32a^2b+8a^2c+20a^2+88ab+54a+41)}{\sum\limits_{cyc}(32a^2b+16a^2c+24a^2+84ab+54a+33)}\cdot\frac{\sum\limits_{cyc}(4ab+a+4)}{\sum\limits_{cyc}(4ab+2a+3)}\leq1$$ or $$\frac{\sum\limits_{cyc}(32a^2b+8a^2c+20a^2+88ab+54a+41)}{\sum\limits_{cyc}(32a^2b+16a^2c+24a^2+84ab+54a+33)}-1\leq\frac{\sum\limits_{cyc}(4ab+2a+3)}{\sum\limits_{cyc}(4ab+a+4)}-1$$ or $$\sum_{cyc}(a-1)\sum\limits_{cyc}(32a^2b+16a^2c+24a^2+84ab+54a+33)+$$ $$+\sum_{cyc}(8a^2c-8+4a^2-4ab)\sum\limits_{cyc}(4ab+a+4)\geq0,$$ which is obvious by AM-GM: $$\sum_{cyc}(a-1)=a+b+c-3\geq3\sqrt[3]{abc}-3=0$$ and $$\sum_{cyc}(8a^2c-8+4a^2-4ab)=$$ $$=8(a^2c+b^2a+c^2b-3)+2\sum_{cyc}(a^2+b^2-2ab)\geq$$ $$\geq8\left(3\sqrt[3]{a^2c\cdot b^2a\cdot c^2b}-3\right)+2\sum_{cyc}\left(2\sqrt{a^2b^2}-2ab\right)=0.$$

Answer 1

With the substitutions $a = \frac{x}{y}, \ b = \frac{y}{z}$ and $c = \frac{z}{x}$, it suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le 1.$$ It suffices to prove that (The desired result follows by summing cyclically.) $$\sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le \frac{9x^2+9z^2 + 16xy + 4yz + 34zx}{18x^2+18y^2+18z^2+54xy+54yz+54zx}.$$ Squaring both sides, it suffices to prove that $f(x, y, z) \ge 0$ where \begin{align} f(x,y,z) &= 162 x^4 y^2-549 x^4 y z+504 x^4 z^2+576 x^3 y^3-452 x^3 y^2 z-1300 x^3 y z^2+1708 x^3 z^3+512 x^2 y^4\nonumber\\ &\quad +1144 x^2 y^3 z-1148 x^2 y^2 z^2-1582 x^2 y z^3+504 x^2 z^4-68 x y^4 z-440 x y^3 z^2-596 x y^2 z^3\nonumber\\ &\quad+180 x y z^4+32 y^4 z^2+192 y^3 z^3+378 y^2 z^4+243 y z^5. \end{align} We use the Buffalo Way. There are three possible cases:

1) $z = \min(x,y,z)$: Let $y = z + s, \ x = z+ t; \ s, t\ge 0$. We have $$f(z+t, z+s, z) = a_4z^4 + a_3z^3 + a_2z^2 + a_1z + a_0$$ where \begin{align} a_4 &= 5616 s^2-1728 s t+1728 t^2, \\ a_3 &= 3376 s^3+12864 s^2 t-1176 s t^2+1000 t^3, \\ a_2 &= 476 s^4+7400 s^3 t+10156 s^2 t^2-1376 s t^3+117 t^4, \\ a_1 &= 956 s^4 t+4920 s^3 t^2+1924 s^2 t^3-225 s t^4, \\ a_0 &= 512 s^4 t^2+576 s^3 t^3+162 s^2 t^4. \end{align} It is easy to prove that $a_4\ge 0, \ a_3\ge 0, \ a_2 \ge 0, \ a_0 \ge 0$ and $4a_2a_0 \ge a_1^2$. Thus, $f(z+t, z+s, z) \ge 0$.

2) $y = \min(x,y,z)$: Let $z = y+s, \ x = y+t; \ s, t\ge 0$. We have \begin{align} f(y+t, y, y+s) &= (5616 s^2-1728 s t+1728 t^2) y^4+(6400 s^3+6600 s^2 t+5088 s t^2+1000 t^3) y^3\nonumber\\ &\quad +(2277 s^4+6116 s^3 t+11626 s^2 t^2+3908 s t^3+117 t^4) y^2\nonumber\\ &\quad +(243 s^5+1188 s^4 t+5558 s^3 t^2+5840 s^2 t^3+459 s t^4) y+504 s^4 t^2+1708 s^3 t^3+504 s^2 t^4. \end{align} Clearly, $f(y+t, y, y+s)\ge 0$.

3) $x = \min(x,y,z)$: Similar.

Answer 2

My second proof.

By Cauchy-Bunyakovsky-Schwarz inequality, it suffices to prove that $$\left(\sum_{\mathrm{cyc}}\frac{2a+1}{4a + 2b + 3}\right)\left(\sum_{\mathrm{cyc}}\frac{a}{2a+1}\right) \le 1. \tag{1}$$

We use the pqr method. Let $p := a + b + c, q := ab + bc + ca, r := abc = 1$. We have $p\ge 3$. (1) is written as $$ \left( 32\,p-96 \right) P+ \left( 24\,p+32\,q+48 \right) Q $$ $$+ 28\,{p}^{3 }+16\,{p}^{2}q+30\,{p}^{2}+24\,pq-48\,{q}^{2}-87\,p-348\,q-585 \ge 0 \tag{2}$$ where $P := b/a + c/b + a/c$ and $Q := a/b + b/c + c/a$.

By Cauchy-Bunyakovsky-Schwarz inequality, we have $P \ge \frac{(a+b+c)^2}{ab + bc + ca}$ and $Q \ge \frac{(a+b+c)^2}{ab + bc + ca}$. Using (2) and $P \ge p^2/q, Q \ge p^2/q$, it suffices to prove that $$-48q^3+(16p^2+24p-348)q^2+(28p^3+62p^2-87p-585)q+56p^3-48p^2 \ge 0. \tag{3}$$

Let $t := p - 3 \ge 0$ and $s := p^2/q - 3 \ge 0$, (3) can be written as $\frac{(3+t)^2}{(3+s)^3}f(s, t) \ge 0$ where $$f(s, t) := 28\,{s}^{2}{t}^{3}+16\,s{t}^{4}+56\,{s}^{3}t+314\,{s}^{2}{t}^{2}+384\, s{t}^{3}+120\,{s}^{3}+1545\,{s}^{2}t$$ $$+2616\,s{t}^{2}+324\,{t}^{3}+1548 \,{s}^{2}+8046\,st+2430\,{t}^{2}+4860\,s+6561\,t. $$ We are done.

Answer 3

We begin with a theorem :

Theorem :

Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :

$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :

$$a+b+c\geq d+e+f$$

Here we suppose that we have :

$a\geq b \geq 1 \geq c $

So to get the majorization we prove this :

$\sqrt{\frac{a}{4a+2b+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Wich is equivalent to :

$\frac{a}{4a+2b+3}\geq \frac{c}{4c+2a+3}$

Or :

$a(4c+2a+3)\geq c(4a+2b+3)$

Wich is obvious under the previous conditions.

With the same reasoning we can prove that we have :

$\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{a}{4a+2b+3}}\geq\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{8.2a}\geq 0.5-\frac{1}{8.2b}\geq \frac{1}{8.2a}+\frac{1}{8.2b} $ wich corresponding to $d\geq e \geq f $ in the initial theorem

It's clear that we have :

$\sqrt{\frac{a}{4a+2b+3}}\leq \sqrt{\frac{a}{4a+3}}\leq 0.5-\frac{1}{8.2a}$

And

$\sqrt{\frac{b}{4b+2c+3}}\leq \sqrt{\frac{b}{4b+3}}\leq 0.5-\frac{1}{8.2b}$

So we have :

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})$

And

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\sqrt{\frac{c}{4c+2a+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})(\frac{1}{8.2a}+\frac{1}{8.2b})$

Wich is true because we have with the condition $abc=1$

$$27\leq \prod_{cyc}\sqrt{4a+2b+3}$$

So now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{a}{4a+2b+3}} \geq \sqrt{\frac{c}{4c+2a+3}}$ is the same.

And for the case $a\geq 1 \geq b \geq c$ you just have to make the following substitution $B=\frac{1}{b}$ to find the previous case $a\geq b \geq 1 \geq c $

Edit :

With the previous substitution the original inequality becomes with $a\geq b \geq 1 \geq c$ and $ac=b$:

$$\sqrt{\frac{ab}{4ab+2+3b}}+\sqrt{\frac{1}{4+2cb+3b}}+\sqrt{\frac{c}{4c+2a+3}}$$

We can briefly prove that we have :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}$

And

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{11(ab)^2}\geq \frac{1}{3} \geq 1-\frac{1}{3}-(0.5-\frac{1}{11(ab)^2})$

wich corresponding to $d\geq e \geq f $ in the initial theorem

Now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}\geq \sqrt{\frac{1}{4+2cb+3b}}$ works this the same majorization.