Finding minimum $\sqrt{a(2b+1)}+\sqrt{b(2c+1)}+\sqrt{c(2a+1)}$?

Question

Recently, my friend has sent me an excessively hard inequality problem, it's

If $a,b,c\ge 0: ab+bc+ca=3$ then find the minimum $$P=\sqrt{a(2b+1)}+\sqrt{b(2c+1)}+\sqrt{c(2a+1)}.$$

I try to test some specific value.

$a=b=c=1$ then $P=3\sqrt{3}\approx 5.1962$

$a=b=\sqrt{3}; c=0$ then $P\approx 4.0967:= P_{0}$

Hence, we will prove $$\sqrt{a(2b+1)}+\sqrt{b(2c+1)}+\sqrt{c(2a+1)}\ge P_{0}.$$ After squaring both side and use the hypothesis we obtain $$a+b+c+2\sum_{cyc}\sqrt{ab(2b+1)(2c+1)}\ge P^2_{0}-6.$$ I don't know how to prove the last inequality.

Could you help me make it clear two question?

1. Is $P_{0}$ good enough to be the desired minimum?

2. Is there some popular ways for the square root cyclic inequality?

Thank you for your help. Any ideas and comments are welcome and appreciate.

Update $1$.

As Michael Rozenberg pointed out, I checked $$\sqrt{a(2b+1)}+\sqrt{b(2c+1)}+\sqrt{c(2a+1)}\ge 4.$$ Equality occurs iff $a=3;b=1;c=0$ and and its cyclic permutations.

Update $2$.

It seems that the following inequality is true.

$$\color{black}{\sqrt{ab+ka}+\sqrt{bc+kb}+\sqrt{ca+kc}\ge \sqrt{q+qk}+\sqrt{k}}$$ holds for all $a,b,c\ge 0$ satisfying $q=ab+bc+ca>1; k=\dfrac{1}{q-1}.$

Equality holds at $(a,b,c)=(q,1,0)$ and its cyclic permutations.

For $k=2,$ see also here. I hope the proof in link is useful for general cases.

Answer 1

This solution is motivated by the first approach by River Li. Our goal is to show the below.

$$\sum_{cyc}ab=3 \quad \Rightarrow \quad \sum_{cyc} \sqrt{a(2b+1)} \geq 4 $$

Step 1

Squaring both sides and expanding, it suffices to show the below.

$$2\sum_{cyc} \sqrt{ab(2b+1)(2c+1)} \geq 10-\sum_{cyc} a$$

Now, by the first approach by River Li, it suffices to show the below.

$$2\sqrt{ \sum_{cyc}ab(2b+1)(2c+1)} \geq 10-\sum_{cyc} a$$

Or, by squaring and expanding, it suffices to show the below when $\sum_{cyc} a \leq 10$.

$$16abc (\sum_{cyc} a) + 24abc + 8 \sum_{cyc} a b^2 + 4 \sum_{cyc} ab \geq 100-20\sum_{cyc} a + (\sum_{cyc} a)^2$$

Step 2

Our crucial idea is to symmetrize the LHS. It is easy to think the following by using the equality condition.

$$\sum_{cyc} a(3b+c-3)^2 \geq 0$$

$$(\sum_{cyc} a) (\sum_{cyc} ab)= 3 \sum_{cyc} a$$

Subtracting these two, and by expanding, we get the following.

$$16abc (\sum_{cyc} a) + 24abc + 8 \sum_{cyc} a b^2 + 4 \sum_{cyc} ab$$

$$= \sum_{cyc} a (3b+c-3)^2 -12 \sum_{cyc} a + 84 + 9abc + 16abc (\sum_{cyc} a) $$

$$\geq \sum_{cyc} a (3b+c-3)^2 -12 \sum_{cyc} a + 84$$

Therefore, by factorizing, it suffices to show the below.

$$\sum_{cyc} a (3b+c-3)^2 \geq (\sum_{cyc} a -4)^2$$

Step 3

Note that by C-S inequality, the following holds.

$$\bigg( \sum_{cyc} a \bigg) \bigg( \sum_{cyc} a (3b+c-3)^2 \bigg) \geq \bigg( \sum_{cyc} a (3b+c-3) \bigg)^2 $$

$$= (4\sum_{cyc} ab -3\sum_{cyc} a)^2 = 9(\sum_{cyc} a -4)^2 $$

Therefore, whenever $\sum_{cyc} a \leq 9$, the following holds.

$$\sum_{cyc} a (3b+c-3)^2 \geq (\sum_{cyc} a -4)^2$$

Step 4

Our final step is to consider the case when $\sum_{cyc} a > 9$. In this case, it is trivial that the following holds.

$$2\sqrt{ \sum_{cyc}ab(2b+1)(2c+1)} \geq 2\sqrt{ \sum_{cyc}ab} $$

$$ = 2\sqrt{3} > 10-\sum_{cyc} a$$

Now the problem is solved. The equality holds if and only if $(a,b,c)=(3,1,0),(1,0,3),(0,3,1)$.

Answer 2

Here are two approaches.

Approach 1.

Let $$x := a(2b + 1), \quad y := b(2c + 1), \quad z := c(2a + 1).$$

We need to prove that $$\sqrt{x} + \sqrt{y} + \sqrt{z} \ge 4. \tag{1}$$

Squaring both sides of (1), it suffices to prove that $$x + y + z + 2\sqrt{xy} + 2\sqrt{yz} + 2\sqrt{zx} \ge 16.$$

Using $\sqrt{xy} + \sqrt{yz} + \sqrt{zx} \ge \sqrt{xy + yz + zx}$ (easy), it suffices to prove that $$x + y + z + 2\sqrt{xy + yz + zx} \ge 16. \tag{2}$$

(2) is true which is verified by Mathematica.

$\phantom{2}$

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Approach 2.

WLOG, assume that $c = \min(a, b, c)$.

Let $$x := a(2b + 1), \quad y := b(2c + 1), \quad z := c(2a + 1).$$

Using $\sqrt{y} + \sqrt{z} \ge \sqrt{y + z}$ (easy), it suffices to prove that $$\sqrt{x} + \sqrt{y + z} \ge 4$$ or (squaring both sides) $$x + y + z + 2\sqrt{x(y + z)} \ge 16$$ or $$x + y + z + 6 \cdot \frac{2 x(y + z) }{2\sqrt{x(y + z)\cdot 9}} \ge 16$$ or (by AM-GM) $$x + y + z + 6 \cdot \frac{2x(y + z)}{x(y + z) + 9} \ge 16. \tag{1}$$

(1) is true which is verified by Mathematica. I have a proof below.

Using $c = \frac{3 - ab}{a + b}$, (1) is equivalently written as \begin{align*} &-4\,{a}^{5}{b}^{2}-8\,{a}^{4}{b}^{3}-8\,{a}^{3}{b}^{4}-4\,{a}^{2}{b}^{ 5}-2\,{a}^{5}b-12\,{a}^{4}{b}^{2}-18\,{a}^{3}{b}^{3}\\ &\qquad -8\,{a}^{2}{b}^{4} +2\,a{b}^{5}+8\,{a}^{4}b+5\,{a}^{3}{b}^{2}+13\,{a}^{2}{b}^{3}+17\,a{b} ^{4}+6\,{a}^{4}\\ &\qquad +36\,{a}^{3}b+62\,{a}^{2}{b}^{2}+44\,a{b}^{3}+24\,{a}^{ 3}+93\,{a}^{2}b+84\,a{b}^{2}+9\,{b}^{3}\\ &\qquad -66\,{a}^{2}-138\,ab-90\,{b}^{2 }+36\,a+27\,b\\ &\ge 0. \tag{2} \end{align*}

We can prove that (2) is true for all $a, b > 0$ with $1 \le ab \le 3$.

If $ab = 1$, it is easy to prove that (2) is true.

If $1 < ab \le 3$, we use the substitution $$b = \frac{1}{a} \cdot \frac{s}{1 + s} + \frac{3}{a} \cdot \frac{1}{1 + s}, \quad s \ge 0.$$ (2) is equivalent to $$m_5s^5 + m_4s^4 + m_3s^3 + m_2s^2 + m_1s + m_0 \ge 0$$ where $m_5, \cdots, m_0$ are polynomials in $a$. It is easy to prove that $m_5, \cdots, m_0 \ge 0$.

We are done.

Answer 3

Another idea.

WLOG, assume $a=\max(a,b,c)$, let's symmetrize the LHS in terms of $b,c$.
We will prove that: \begin{align*} \sqrt{a+2ab} +\sqrt{b+2bc} + \sqrt{c+2ca} \ge \sqrt{a+2ab+2ac} + \sqrt{b+2bc+c}\end{align*} By squaring: \begin{align*} \sum_{\text{cyc}} \sqrt{\left(a+2ab \right)\left(b+2bc \right)} \ge \sqrt{\left(a+2ab+2ac \right) \left(b+2bc+c \right)}\end{align*} Which is true, because: \begin{align} & \sqrt{\left(a+2ab \right)\left(b+2bc \right)} + \sqrt{\left(b+2bc \right)\left(c+2ca \right)} + \sqrt{\left(c+2ca \right)\left(a+2ab \right)} \\ & \ge \sqrt{\left(a+2ab \right)\left(b+2bc \right)} + \sqrt{\left(b+2bc \right) \cdot 2ca} + \sqrt{c \left( a+2ab \right) + 2c^2a} \\ & \ge \sqrt{\left(a+2ab \right)\left(b+2bc \right) + \left(b+2bc \right) \cdot 2ca + c \left( a+2ab \right) + 2c^2a} \\ & = \sqrt{\left(a+2ab+2ac \right) \left(b+2bc+c \right)} \end{align} It's enough to prove: \begin{align*} \sqrt{a+2ab+2ac} + \sqrt{b+2bc+c} \ge 4\end{align*}

Let $b+c=s$ and $bc=p$, where $s \ge 0, s\ge 2\sqrt{p}, p \in [0, 1]$, implies $a=\frac{3-bc}{b+c}=\frac{3-p}{s}$.
The above inequality is equivalent with: \begin{align*} \sqrt{\left(2 + \frac{1}{s} \right) \left(3-p \right)} + \sqrt{s+2p} \ge 4\end{align*} We will prove that this inequality holds true for all $s>0$ and $p \in [0, 1]$. Obvisously, $\text{LHS}$ is a concave function of $p$, since $f(x) = \sqrt{1+ax}$ is concave for all $a \neq 0$. Thus it suffices to prove the inequality in cases $p=0$ and $p=1$.

• If $p=0$, then: \begin{align*} \text{LHS} = \sqrt{6+\frac{3}{s}} + \sqrt{s} = \sqrt{3\left(1+1+\frac{1}{s} \right)} +\sqrt{s} \ge \frac{3}{\sqrt[6]{s}} +\sqrt{s} \ge 4\end{align*} by AM-GM inequality.
• If $p=1 $, then: \begin{align*} \text{LHS} = \sqrt{4+\frac{2}{s}} + \sqrt{s+2}& =\sqrt{2+2+\frac{2}{s}} + \sqrt{s+1+1} \\ &\ge \sqrt{\frac{6}{\sqrt[3]{s}}} +\sqrt{3\sqrt[3]{s}} \\ & \ge 2\sqrt[4]{18} \\ &>4\end{align*} by AM-GM inequality.

The proof is completed.