Prove that $\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$

Question

Let $a$, $b$ and $c$ be non-negative numbers. Prove that: $$\sum\limits_{cyc}\sqrt{a^2+10bc}\geq\frac{1}{2}\sum\limits_{cyc}\sqrt{22a(b+c)}$$ I tried SOS, C-S, Holder, Mixing Variables and more, but without success.

For example, after squaring of the both sides we need to prove $$\sum_{cyc}\left(a^2+10ab+2\sqrt{(a^2+10bc)(b^2+10ac)}\right)\geq$$ $$\geq\frac{11}{2}\sum_{cyc}\left(2ab+2\sqrt{ab(b+c)(a+c))}\right)$$ or $$\sum_{cyc}\left(3a^2+30ab-\left(\sqrt{a^2+10bc}-\sqrt{b^2+10ac}\right)^2\right)\geq$$ $$\geq\frac{11}{2}\sum_{cyc}\left(6ab-\left(\sqrt{a(b+c)}-\sqrt{b(a+c)}\right)^2\right)$$ or $$\sum_{cyc}(3a^2-3ab)-\sum_{cyc}\frac{(a-b)^2(a+b-10c)^2}{\left(\sqrt{a^2+10bc}+\sqrt{b^2+10ac}\right)^2}+$$ $$+\frac{11}{2}\sum_{cyc}\frac{(a-b)^2c^2}{\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)^2}\geq0$$ or $$\sum_{cyc}(a-b)^2\left(3-\frac{2(a+b-10c)^2}{\left(\sqrt{a^2+10bc}+\sqrt{b^2+10ac}\right)^2}+\frac{11c^2}{\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)^2}\right)\geq0$$ and from here I don't see, how to continue.

Thank you for your help!

Answer 1

The Buffalo Way works although it is an ugly solution.

WLOG, assume that $a+b+c=3$. Write the inequality as $$\sum_{\mathrm{cyc}} \sqrt{\frac{a^2+10bc}{11}} \ge \sum_{\mathrm{cyc}} \sqrt{\frac{a(b+c)}{2}}.$$

Let $X = \frac{a^2+10bc}{11}, \ Y = \frac{b^2+10ca}{11}, \ Z = \frac{c^2+10ab}{11}$ and $U = \frac{a(b+c)}{2}, \ V = \frac{b(c+a)}{2}, \ W = \frac{c(a+b)}{2}$.

We will use the following bounds: $$\sqrt{x} \le \frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}, \quad \forall x > 0$$ and $$\sqrt{x} \ge \frac{2(x+3)(3x+1)x}{(x+1)(x^2+14x+1)}, \quad \forall x > 0$$ which follows from $$\Big(\frac{(x+1)(x^2+14x+1)}{2(x+3)(3x+1)}\Big)^2 - x = \frac{(x-1)^6}{4(x+3)^2(3x+1)^2}.$$

With the bounds above, it suffices to prove that $$\sum_{\mathrm{cyc}(X,Y,Z)} \frac{2(X+3)(3X+1)X}{(X+1)(X^2+14X+1)} \ge \sum_{\mathrm{cyc}(U,V,W)}\frac{(U+1)(U^2+14U+1)}{2(U+3)(3U+1)}.$$

After homogenization, it suffices to prove that $f(a,b,c)\ge 0$ where $f(a,b,c)$ is a homogeneous polynomial with degree $32$.

We use the Buffalo Way. Due to symmetry, assume that $c\le b\le a$.

If $c = 0$, $f(a,b,0)$ is a polynomial with non-negative coefficients. True.

If $c > 0$, let $c=1, \ b = 1+s, \ a = 1+s+t; \ s,t\ge 0$. $f(1+s+t, 1+s, 1)$ is a polynomial with non-negative coefficients. True.

We are done.

Answer 2

Disclaimer: It is a very ugly solution. All of the coefficients below are obtained from a lot of trial and error.

By Holder, $$ (\sum \sqrt{a^2+10bc})^2 \ge \frac{\left(\sum (a^2+10bc)(2a^2+2b^2+2c^2+4bc+21ab+21ac) \right)^3}{\sum (a^2+10bc)^2(2a^2+2b^2+2c^2+4bc+21ab+21ac)^3} $$ By Cauchy-Schwarz, $$ \frac{11}{2} \left(\sum \sqrt{ab+ac}\right)^2 \leq \frac{11}{2} \left( \sum a^2+b^2+c^2+11bc+50ab+50ac \right)\left( \sum \frac{ab+ac}{a^2+b^2+c^2+11bc+50ab+50ac} \right) $$

Now it remains to prove that $$ \frac{\left(\sum (a^2+10bc)(2a^2+2b^2+2c^2+4bc+21ab+21ac) \right)^3}{\sum (a^2+10bc)^2(2a^2+2b^2+2c^2+4bc+21ab+21ac)^3} \ge \frac{11}{2} \left( \sum a^2+b^2+c^2+11bc+50ab+50ac \right)\left( \sum \frac{ab+ac}{a^2+b^2+c^2+11bc+50ab+50ac} \right) $$

Here is a plot of $LHS-RHS$ assuming $a+b+c = 3$:

After expanding, we need to prove the nonnegativity of a symmetric polynomial of degree 18.

B-W has a few negative coefficients. But the SOS solver at

https://aistudio.baidu.com/application/detail/37245

gives a solution. It is so long that it exceeds the StackExchange length limit, so I cannot include it here.

Interesting but Unsuccessful Attempt: A simpler Cauchy-Schwarz step is $$ \frac{11}{2} \left(\sum \sqrt{ab+ac}\right)^2 \leq \frac{11}{2} \left( \sum 19a+5b+5c \right)\left( \sum \frac{ab+ac}{19a+5b+5c} \right) $$ And $$\frac{11}{2} \left( \sum 19a+5b+5c \right)\left( \sum \frac{ab+ac}{19a+5b+5c} \right) \le (\sum \sqrt{a^2+10bc})^2$$ is numerically correct. However, I cannot find a Holder step that works here.

Answer 3

Remark. The SOS (Sum of Squares) method works. The proof of $S_c \ge 0$ is simple. The proof of $S_b \ge 0$ is a little bit complicated. The proof of $a^2S_b + b^2S_a \ge 0$ is very complicated. Hope to see better proof of $a^2S_b + b^2S_a\ge 0$.

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

According to the OP's result, it suffices to prove that $$S_c(a-b)^2 + S_a(b-c)^2 + S_b(c-a)^2 \ge 0$$ where $$S_c = 3-\frac{2(a+b-10c)^2}{\left(\sqrt{a^2+10bc}+\sqrt{b^2+10ac}\right)^2}+\frac{11c^2}{\left(\sqrt{a(b+c)}+\sqrt{b(a+c)}\right)^2}, $$ $$S_a = 3-\frac{2(b+c-10a)^2}{\left(\sqrt{b^2+10ca}+\sqrt{c^2+10ba}\right)^2}+\frac{11a^2}{\left(\sqrt{b(c+a)}+\sqrt{c(b+a)}\right)^2}, $$ $$S_b = 3-\frac{2(c+a-10b)^2}{\left(\sqrt{c^2+10ab}+\sqrt{a^2+10cb}\right)^2}+\frac{11b^2}{\left(\sqrt{c(a+b)}+\sqrt{a(c+b)}\right)^2}. $$

According to the SOS method, it suffices to prove that $$S_c \ge 0, \quad S_b \ge 0, \quad a^2 S_b + b^2 S_a \ge 0.$$

• First, we have \begin{align*} S_c &\ge 3-\frac{2(a+b-10c)^2}{\left(\sqrt{a^2+10bc}+\sqrt{b^2+10ac}\right)^2}\\ &= 3-\frac{2(a+b-10c)^2}{a^2+10bc + b^2+10ac + 2\sqrt{(a^2+10bc)(b^2+10ac)}}\\ &\ge 3-\frac{2(a+b-10c)^2}{a^2+10bc + b^2+10ac + 2(ab + 10c\sqrt{ab})}\\ &= 3-\frac{2(a+b-10c)^2}{(a+b)^2 + 10c(a+b) + 20c\sqrt{ab}}\\ &\ge 3-\frac{2(a+b-10c)^2}{(a+b)^2 + 10c(a+b) + 20c\sqrt{c(a+b-c)}}\\ &\ge 0 \end{align*} where we use Cauchy-Bunyakovsky-Schwarz inequality, and $ab - c(a+b-c) = (a-c)(b-c) \ge 0$. For the last inequality: Letting $c = 1$ and $x := a + b \ge 2$, it suffices to prove that $3-\frac{2(x-10)^2}{x^2 + 10x + 20\sqrt{x-1}} \ge 0$ (easy).

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• Second, let us prove that $S_b \ge 0$. If $a = c$, true. If $a > c$, we have $$ S_b = 3 - \frac{2(\sqrt{c^2+10ab} - \sqrt{a^2+10bc})^2}{(a-c)^2} + \frac{11(\sqrt{c(b+a)} - \sqrt{a(b+c)})^2}{(a-c)^2}.$$ It suffices to prove that $$a^2 + c^2 + 16ac - 9ab - 9bc + 4\sqrt{(c^2+10ab)(a^2+10bc)} \ge 22\sqrt{c(b+a)a(b+c)},$$ or (using $a^2 + c^2 + 16ac - 9ab - 9bc + 4\sqrt{(c^2+10ab)(a^2+10bc)}\ge 0$) $$\left(a^2 + c^2 + 16ac - 9ab - 9bc + 4\sqrt{(c^2+10ab)(a^2+10bc)}\right)^2 $$ $$\ge \left(22\sqrt{c(b+a)a(b+c)}\right)^2,$$ or $$(8a^2-72ab+128ac-72bc+8c^2)\sqrt{(c^2+10ab)(a^2+10bc)}$$ $$+a^4+142a^3b+32a^3c+81a^2b^2-790a^2bc-210a^2c^2+1278ab^2c$$ $$-790abc^2+32ac^3+81b^2c^2+142bc^3+c^4\ge 0. \tag{1}$$ (1) is true which is verified by Mathematica - a Computer Algebra System (CAS). Is there a nice proof of (1)?

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• Third, we have \begin{align*} &a^2S_b + b^2S_a \\ ={}& 3a^2 + 3b^2 - \frac{2(c+a-10b)^2a^2}{\left(\sqrt{c^2+10ab}+\sqrt{a^2+10cb}\right)^2} -\frac{2(b+c-10a)^2b^2}{\left(\sqrt{b^2+10ca}+\sqrt{c^2+10ba}\right)^2}\\ &\qquad +\frac{11b^2a^2}{\left(\sqrt{c(a+b)}+\sqrt{a(c+b)}\right)^2}+\frac{11a^2b^2}{\left(\sqrt{b(c+a)}+\sqrt{c(b+a)}\right)^2}\\ \ge{}& 3a^2 + 3b^2 - \frac{2(c+a-10b)^2a^2}{\left(\sqrt{c^2+10ab}+\sqrt{a^2+10cb}\right)^2} -\frac{2(b+c-10a)^2b^2}{\left(\sqrt{b^2+10ca}+\sqrt{c^2+10ba}\right)^2}\\ &\qquad +2 \times\frac{11b^2a^2}{\left(\sqrt{c(a+b)}+\sqrt{a(c+b)}\right)^2}\\ \ge{}& 3a^2 + 3b^2 - \frac{2(c+a-10b)^2a^2}{c^2+10ab+a^2+10cb + 10a^2\cdot \frac{2A}{1+A}} -\frac{2(b+c-10a)^2b^2}{b^2+10ca+c^2+10ba+10a^2\cdot \frac{2C}{1+C}}\\ &\qquad +2 \times\frac{11b^2a^2}{c(a+b) + a(c+b) + 2\left(\frac{B^2}{2} + \frac{c(a+b)a(c+b)}{2B^2}\right)}\\ \ge{}& 0 \tag{1} \end{align*} where $A = \frac{(c^2+10ab)(a^2+10cb)}{25a^4}$ and $C = \frac{(b^2+10ca)(c^2+10ba)}{25a^4}$ and $B = \frac{a+b+c}{3}$, and we use $(\sqrt{c(a+b)}+\sqrt{a(c+b)})^2 \ge (\sqrt{b(c+a)}+\sqrt{c(b+a)})^2$. The last inequality can be proved by letting $a = c + s + t, b = c + s$ for $s, t \ge 0$ (all coefficients of the resulting polynomial are non-negative).

Remark. We can also use $$S_b = 3 - \frac{2(\sqrt{c^2+10ab} - \sqrt{a^2+10bc})^2}{(a-c)^2} + \frac{11(\sqrt{c(b+a)} - \sqrt{a(b+c)})^2}{(a-c)^2},$$ and $$S_a = 3 - \frac{2(\sqrt{b^2+10ca} - \sqrt{c^2+10ab})^2}{(b-c)^2} + \frac{11(\sqrt{b(a+c)} - \sqrt{c(b+a)})^2}{(b-c)^2}.$$ But I have no idea how to proceed.