# Prove that $\sum\limits_{cyc}\sqrt{a+11bc+6}\geq9\sqrt2.$

Let $$a$$, $$b$$ and $$c$$ be non-negative numbers such that $$ab+ac+bc+abc=4.$$ Prove that: $$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}+\sqrt{c+11ab+6}\geq 9\sqrt2.$$

The equality occurs for $$(a,b,c)=(1,1,1)$$ and again for $$(a,b,c)=(2,2,0)$$ and for the cyclic permutations of the last.

I tried Holder: $$\sum_{cyc}\sqrt{a+11bc+6}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+11bc+6}\right)^2\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}\geq$$ $$\geq \sqrt{\frac{\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3}{\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3}}$$ and it's enough to prove that $$\left(\sum\limits_{cyc}(a+11bc+6)(3a+4)\right)^3\geq162\sum\limits_{cyc}(a+11bc+6)^2(3a+4)^3,$$ which is true for $$(a,b,c)=(2,2,0)$$, $$(a,b,c)=(1,1,1)$$, but it's wrong for $$(a,b,c)=\left(8,\frac{1}{2},0\right).$$

Thanks to River Li for this counterexample.

Also, I tried a substitution $$a=\frac{2x}{y+z},$$ $$b=\frac{2y}{x+z}$$, $$c=\frac{2z}{x+y}$$ and SOS, but it seems very complicated.

Also, I tried the following estimation. By Minkowcki: $$\sqrt{a+11bc+6}+\sqrt{b+11ac+6}\geq\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}.$$

Now, for $$c=\min\{a,b,c\}$$ it's enough to prove that $$\sqrt{(\sqrt{a}+\sqrt{b})^2(1+11c)+24}+\sqrt{c+11ab+6}\geq9\sqrt2,$$ which not so helps.

Also, LM does not help.

Thank you!

Update

Also, there is the following.

We need to prove that: $$\sum_{cyc}\sqrt{\frac{2x}{y+z}+\frac{44yz}{(x+y)(x+z)}+6}\geq9\sqrt2$$ or $$\sum_{cyc}\sqrt{\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3}\geq9,$$ where $$x$$, $$y$$ and $$z$$ are non-negatives such that $$xy+xz+yz\neq0.$$

Now, by Holder $$\left(\sum_{cyc}\sqrt{\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3}\right)^2\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3\geq$$ $$\geq\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3,$$ where $$k$$, $$m$$ and $$n$$ are reals such that the expression $$kx^2+y^2+z^2+myz+nxy+nxz$$

is non-negative for all non-negatives $$x$$, $$y$$ and $$z$$.

Thus, it's enough to choose values of $$k$$, $$m$$ and $$n$$ for which the following inequality is true. $$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(kx^2+y^2+z^2+myz+nxy+nxz)\right)^3\geq$$ $$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(kx^2+y^2+z^2+myz+nxy+nxz)^3$$ From the equality case we can get that should be $$2k-5m+2n=8.$$ For $$k=1$$, $$m=0$$ and $$n=3$$ we need to prove that:

$$\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\geq$$ $$\geq81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3,$$ which is true for $$y=z$$ and it's true for $$z=0$$, but I have no a proof for all non-negative variables.

• Holder part: I check $a=1/2, b=8, c=0$, negative. Do I miss something? Jun 26, 2019 at 3:19
• @River Li $\sqrt{6.5}+\sqrt{14}+\sqrt{50}>9\sqrt2.$ Jun 26, 2019 at 3:22
• I mean the inequality before "which I don't know how to prove and I don't see a counterexample" Jun 26, 2019 at 3:34
• @River Li Wow! Thank you very much! Which says that this way is wrong. Jun 26, 2019 at 3:40

Michael Rozenberg actually gave a proof. I do a little bit by the Buffalo Way to prove that \begin{align} &\left(\sum_{cyc}\left(\frac{x}{y+z}+\frac{22yz}{(x+y)(x+z)}+3\right)(x^2+y^2+z^2+3xy+3xz)\right)^3\\ \geq\ & 81\sum_{cyc}\left(\tfrac{x}{y+z}+\tfrac{22yz}{(x+y)(x+z)}+3\right)^2(x^2+y^2+z^2+3xy+3xz)^3. \end{align}
It suffices to prove that $$f(x,y,z)\ge 0$$ where $$f(x,y,z)$$ is a polynomial (a long expression).
WLOG, assume that $$z = \min(x,y,z).$$ There are two possible cases:
1) If $$z \le y \le x$$, let $$y=z+s, \ x = z+s+t; \ s,t \ge 0$$. Note that $$f(z+s+t, z+s, z)$$ is a polynomial in $$z, s, t$$ with non-negative coefficients. It is true.
2) If $$z \le x\le y$$, let $$x = z+s, \ y = z+s+t; \ s,t \ge 0.$$ Note that $$f(z+s, z+s+t, z)$$ is a polynomial in $$z, s, t$$ with non-negative coefficients. It is true. We are done.
• How did you make it? Did you use some software? By the way, it's enough to consider one case only: $x\leq y\leq z$. Jun 29, 2019 at 4:06