Proving $\sum\limits_{cyc} \dfrac{1}{\sqrt{2a+bc}} \ge 1+\sqrt{2}$ for $a,b,c \ge 0$ and $ab+bc+ca=1$

Question

Let $a,b,c \ge 0 : ab+bc+ca=1.$ Prove that $$\dfrac{1}{\sqrt{2a+bc}}+\dfrac{1}{\sqrt{2b+ca}}+\dfrac{1}{\sqrt{2c+ab}} \ge 1+\sqrt{2}.$$

Source: here. and [here] If $a,b,c\ge 0: ab+bc+ca=1$, find Min $P=\frac{1}{\sqrt{2a+bc}}+\frac{1}{\sqrt{2b+ca}}+\frac{1}{\sqrt{2c+ab}}$.

I tried Holder, or AM-GM, Cauchy - Schwarz but it is not good.

And this is an old Problem, it's easier.

Let $a,b,c \ge 0 : ab+bc+ca=1.$ Prove that $$\dfrac{1}{2a+bc}+\dfrac{1}{2b+ca}+\dfrac{1}{2c+ab} \ge 2.$$

By AM-GM $$\sum\limits_{cyc}\dfrac{1}{2a+bc} \ge \sum\limits_{cyc}\dfrac{1}{\dfrac{a}{b+c}+a(b+c)+ca} =\sum\limits_{cyc}\dfrac{b+c}{a+b+c}=2.$$

I hope this problem will be solved by a nice solution.

Thanks.

Answer 1

My second proof using Holder inequality.

By Holder inequality, we have \begin{align*} &\left(\sum_{\mathrm{cyc}}\frac{1}{\sqrt{2a + bc}}\right)^2 \sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3\\ \ge{}& \left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3. \end{align*}

It suffices to prove that \begin{align*} &\left(6ab + 6bc + 6ca + 2a\sqrt 2 + 2b\sqrt{2} + 2c\sqrt{2}\right)^3\\ \ge{}& (1 + \sqrt 2)^2\sum_{\mathrm{cyc}} (2a + bc)(2ab + 2bc + 2ca + b\sqrt 2 + c\sqrt 2)^3. \tag{1} \end{align*}

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc$. Using $p^2 \ge 3q = 3$, we have $p \ge \sqrt 3$.

(1) is written as $$c_1 r + c_0 \ge 0\tag{2}$$ where \begin{align*} c_1 &= 42\,\sqrt {2}{p}^{2}+60\,\sqrt {2}p+56\,{p}^{2}-48\,\sqrt {2}+ 100\,p-88, \\ c_0 &= 10\,\sqrt {2}{p}^{3}-36\,\sqrt {2}{p}^{2}-8\,{p}^{ 3}+100\,\sqrt {2}p+92\,{p}^{2}-136\,\sqrt {2}-168\,p+32. \end{align*} We have $c_1 \ge 0.$

If $p^2 \ge 4q = 4$, we have $c_0 \ge 0$. Thus, (2) is true.

If $p^2 < 4q = 4$, using degree three Schur $r \ge \frac{4pq - p^3}{9} = \frac{4p - p^3}{9}$, we have $$c_1r + c_0 \ge c_1 \cdot \frac{4p - p^3}{9} + c_0 \ge 0.$$

We are done.

Answer 2

Proof.

WLOG, assume that $a \ge b \ge c$.

We split into two cases:

Case 1: $a \ge 2$

By AM-GM, we have $\frac{1}{\sqrt{2a + bc}} = \frac{4}{2\sqrt{(2a + bc)\cdot 4}} \ge \frac{4}{2a + bc + 4}$, and $\frac{1}{\sqrt{2b + ca}} = \frac{2}{2\sqrt{(2b + ca)\cdot 1}} \ge \frac{2}{2b + ca + 1}$, and $\frac{1}{\sqrt{2c + ab}} = \frac{2}{2\sqrt{(2c + ab)\cdot 1}} \ge \frac{2}{2c + ab + 1}$. By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{2}{2b + ca + 1} + \frac{2}{2c + ab + 1} \ge \frac{8}{2b + ca + 1 + 2c + ab + 1}.$$

It suffices to prove that $$\frac{4}{2a + bc + 4} + \frac{8}{2b + ca + 1 + 2c + ab + 1} \ge \frac52 > 1 + \sqrt 2. \tag{1}$$ (1) is true. The proof is given at the end.

$\phantom{2}$

Case 2: $a < 2$

By AM-GM, we have $\frac{1}{\sqrt{2a + bc}} = \frac{2\sqrt 2}{2\sqrt{(2a+bc)\cdot 2}} \ge \frac{2\sqrt 2}{2a + bc + 2}$ and $\frac{1}{\sqrt{2b + ca}} = \frac{2\sqrt 2}{2\sqrt{(2b+ca)\cdot 2}} \ge \frac{2\sqrt 2}{2b + ca + 2}$ and $\frac{1}{\sqrt{2c+ab}} = \frac{2}{2\sqrt{(2c+ab)\cdot 1}} \ge \frac{2}{2c + ab + 1}$.

It suffices to prove that $$\frac{2\sqrt 2}{2a + bc + 2} + \frac{2\sqrt 2}{2b + ca + 2} + \frac{2}{2c + ab + 1} \ge 1 + \sqrt 2$$ or $$\frac{4 - 4ab - abc^2 - 2a^2c - 2b^2c}{(2a + bc + 2)(2b + ca + 2)}\cdot\sqrt 2 + \frac{2}{2c + ab + 1} - 1 \ge 0. \tag{2}$$

Using $b, c \le 1$, we have $4 - 4ab - abc^2 - 2a^2c - 2b^2c \ge 4(bc + ca) - abc - 2a^2 c - 2bc = c(2-a)(2a + b) \ge 0$. Using $\sqrt 2 > 7/5$, it suffices to prove that $$\frac{4 - 4ab - abc^2 - 2a^2c - 2b^2c}{(2a + bc + 2)(2b + ca + 2)}\cdot\frac75 + \frac{2}{2c + ab + 1} - 1 \ge 0. \tag{3}$$ (3) is true. The proof is given at the end.

$\phantom{2}$

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Proof of (1):

Let $p = b + c, q = bc$.

From $ab + bc + ca = 1$, we have $ap + q = 1$ which results in $a = \frac{1 - q}{p}$. From $a \ge 2$, we have $q \le 1 - 2p$.

Thus, we have $$0 < p \le 1/2, \quad 0 \le q \le 1 - 2p. $$

(1) is written as $$f(q) := - \left( 10-5\,p \right) {q}^{2}+ \left( -10\,{p}^{2}+33\,p+8 \right) q+2\, \left( 6\,p+1 \right) \left( 1-2\,p \right)\ge 0. $$

Since $0 < p \le 1/2$, $f(q)$ is concave. We have $f(0) \ge 0$ and $f(1 - 2p) = 10p(1 - 2p)(7 - 2p)\ge 0$. Thus, $f(q) \ge 0$ for all $0 \le q \le 1 - 2p$.

We are done.

$\phantom{2}$

Proof of (3):

Using $c = \frac{1 - ab}{a + b}$, after simplifying (3), it suffices to prove that \begin{align*} &6\,{a}^{4}{b}^{3}+6\,{a}^{3}{b}^{4}-12\,{a}^{5}b-24\,{a}^{4}{b}^{2}-36 \,{a}^{3}{b}^{3}-24\,{a}^{2}{b}^{4}-12\,a{b}^{5}+43\,{a}^{4}b\\ &\quad +76\,{a}^ {3}{b}^{2}+76\,{a}^{2}{b}^{3}+43\,a{b}^{4}+8\,{a}^{4}-2\,{a}^{3}b+4\,{ a}^{2}{b}^{2}-2\,a{b}^{3}+8\,{b}^{4}\\ &\quad -15\,{a}^{3}+2\,{a}^{2}b+2\,a{b}^{ 2}-15\,{b}^{3}-2\,{a}^{2}-16\,ab-2\,{b}^{2}\\ &\ge 0. \tag{4} \end{align*}

From $2 > a \ge b \ge c\ge 0$ and $ab + bc + ca = 1$, we have $$0 < b \le 1, \quad \frac{1}{3b} \le a < 2. \tag{5}$$

Let $$s = \frac{1}{b} - 1, \quad t = \frac{a - \frac{1}{3b}}{2 - a}.$$ Then $s, t \ge 0$ and $$b = \frac{1}{1 + s}, \quad a = \frac{1}{3b}\cdot \frac{1}{1 + t} + \frac{2t}{1 + t}. \tag{6}$$

Using (6), (4) is written as $$\frac{1}{81(1 + t)^5(1 + s)^5}f(s, t)\ge 0$$ where $f(s, t)$ is a polynomial with non-negative coefficients.

We are done.