Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that$$\color{black}{\frac{1}{\sqrt{a+b+7c}}+\frac{1}{\sqrt{c+b+7a}}+\frac{1}{\sqrt{a+c+7b}}\ge 1.}$$
I found the inequality accidentally and there is no original proof. In case it's old problem, I hope there is nice proof like AM-GM, Cauchy-Schwarz,…
Equality holds at $(1,1,1);(0,2,2)$ and that makes some troubles when I tried to use classical inequalities application.
You're welcome to share any ideas and comment here. Thank you for your interest.
Here is what I tried so far.
I thought of Holder inequality but my try is not good enough.
$\bullet$ Holder using 1 $$\left(\sum_{cyc}\frac{1}{b+c+7a}\right)^2\cdot \sum_{cyc}(b+c+7a)(b+c+xa)^3\ge (x+2)^3(a+b+c)^3.$$
Choose $x$ such that equality holds at $a=b=2;c=0.$ Thus, we solve the equation $$256+2\cdot16\cdot(2x+2)^3=64(x+2)^3 \iff x=-2;x=0.$$
Notice that $x=0$ is satisfied. We consider$$\left(\sum_{cyc}\frac{1}{b+c+7a}\right)^2\cdot \sum_{cyc}(b+c+7a)(b+c)^3\ge 8(a+b+c)^3.$$
But $$ 8(a+b+c)^3\ge \sum_{cyc}(b+c+7a)(b+c)^3$$is already wrong at $a=b=\dfrac{9}{10}.$
$\bullet$ Holder using 2 $$\left(\sum_{cyc}\frac{1}{b+c+7a}\right)^2\cdot \sum_{cyc}(b+c+7a)(x+a)^3\ge (a+b+c+3x)^3.$$
Choose $x$ such that equality holds at $a=b=2;c=0.$ Thus, we solve the equation $$4x^3+32(x+2)^3=(4+3x)^3\iff x=-4;x=-\frac{4}{3}. $$
Thus, both Holder using ways are failed. Maybe there is exist a better way.
Best Answer
Proof.
By AM-GM, it suffices to prove that $$\sum_{\mathrm{cyc}} \frac{2}{\frac{a + b + 7c}{2 + c} + (2 + c)} \ge 1. \tag{1}$$
We use the pqr method. Let $p = a + b + c, q = ab + bc + ca, r = abc$.
The condition $a + b + c + abc = 4$ is written as $p + r = 4$. Using $p^3 \ge 27r = 27(4 - p)$, we have $p \ge 3$. Thus, we have $3 \le p \le 4$. Using degree three Schur inequality, we have $r \ge \frac{4pq - p^3}{9}$. Thus, we have $4 - p \ge \frac{4pq - p^3}{9}$ which results in $q \le \frac{p^3 - 9p + 36}{4p}$.
(1) is equivalently written as $$-p{q}^{2}+ \left( -6\,{p}^{2}-44\,p+96 \right) q-{p}^{4}-11\,{p}^{3}+ 75\,{p}^{2}+528\,p-1584 \ge 0. \tag{2}$$
Since $ -6\,{p}^{2}-44\,p+96 < 0$, using $q \le \frac{p^3 - 9p + 36}{4p}$, it suffices to prove that \begin{align*} &-p \cdot \left(\frac{p^3 - 9p + 36}{4p}\right)^2 + \left( -6\,{p}^{2}-44\,p+96 \right) \cdot \frac{p^3 - 9p + 36}{4p}\\[6pt] &\qquad -{p}^{4}-11\,{p}^{3}+ 75\,{p}^{2}+528\,p-1584\\[6pt] &\ge 0 \end{align*} or $$\frac{(p - 3)(4 - p)(p^4 + 47p^3 + 651p^2 + 2265p - 1044)}{16p}\ge 0$$ which is true using $3\le p\le 4$.
We are done.