Prove $\frac{1}{\sqrt{a+b+7c}}+\frac{1}{\sqrt{c+b+7a}}+\frac{1}{\sqrt{a+c+7b}}\ge 1.$

Question

Problem. Given non-negative real numbers $a,b,c$ satisfying $a+b+c+abc=4.$ Prove that$$\color{black}{\frac{1}{\sqrt{a+b+7c}}+\frac{1}{\sqrt{c+b+7a}}+\frac{1}{\sqrt{a+c+7b}}\ge 1.}$$

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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.

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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.

Answer 1

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.

Answer 2

Very complicated for a comment.

1. $abc=0$.

Let $c=0$.

Thus, $a+b=4$ and we need to prove that: $$\frac{1}{\sqrt{a+7b}}+\frac{1}{\sqrt{b+7a}}\geq\frac{1}{2},$$ which is true by Jensen: $$\frac{1}{\sqrt{a+7b}}+\frac{1}{\sqrt{b+7a}}\geq\frac{2}{\sqrt{\frac{a+7b+b+7a}{2}}}=\frac{1}{2}.$$ 2. $abc\neq0$.

Consider $$f(a,b,c)=\sum_{cyc}\frac{1}{\sqrt{a+b+7c}}+\lambda(a+b+c+abc-4)$$ and let $(a,b,c)$ is a minimum point of $f$.

Thus, in this point should be $$\frac{\partial f}{\partial a}=\frac{\partial f}{\partial b}=\frac{\partial f}{\partial c}=0$$ or $$-\frac{1}{2\sqrt{(a+b+7c)^3}}-\frac{1}{2\sqrt{(a+c+7b)^3}}-\frac{7}{2\sqrt{(b+c+7a)^3}}+\lambda(1+bc)=$$ $$=-\frac{1}{2\sqrt{(a+b+7c)^3}}-\frac{7}{2\sqrt{(a+c+7b)^3}}-\frac{1}{2\sqrt{(b+c+7a)^3}}+\lambda(1+ac)=$$$$=-\frac{7}{2\sqrt{(a+b+7c)^3}}-\frac{1}{2\sqrt{(a+c+7b)^3}}-\frac{1}{2\sqrt{(b+c+7a)^3}}+\lambda(1+ab)=0.$$ Now, let in this point $a\neq b$ and $a\neq c$.

Thus, $$\frac{1}{\sqrt{(a+c+7b)^3}}-\frac{1}{\sqrt{(b+c+7a)^3}}=\frac{1}{3}\lambda c(a-b)$$ or $$\frac{(b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2}{c\sqrt{(b+c+7a)^3(a+c+7b)^3}\left(\sqrt{(b+c+7a)^3}+\sqrt{(a+c+7b)^3}\right)}=\frac{1}{18}\lambda.$$ By the similar way we obtain: $$\frac{(b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2}{b\sqrt{(b+c+7a)^3(a+b+7c)^3}\left(\sqrt{(b+c+7a)^3}+\sqrt{(a+b+7c)^3}\right)}=\frac{1}{18}\lambda,$$ which gives $$\frac{(b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2}{c\sqrt{(b+c+7a)^3(a+c+7b)^3}\left(\sqrt{(b+c+7a)^3}+\sqrt{(a+c+7b)^3}\right)}=$$ $$=\frac{(b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2}{b\sqrt{(b+c+7a)^3(a+b+7c)^3}\left(\sqrt{(b+c+7a)^3}+\sqrt{(a+b+7c)^3}\right)}$$ or $$\frac{(b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2}{c\sqrt{(a+c+7b)^3(b+c+7a)^3}+c(a+c+7b)^3}=$$ $$=\frac{(b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2}{b\sqrt{(a+b+7c)^3(b+c+7a)^3}+b(a+b+7c)^3}$$ or $$b(a+b+7c)^3((b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2)-$$ $$-c(a+c+7b)^3((b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2)+\tfrac{\left(b^2(a+b+7c)^3\left((b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2\right)^2-c^2(a+c+7b)^3\left((b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2\right)^2\right)\sqrt{(b+c+7a)^3}}{b\sqrt{(a+b+7c)^3}\left((b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2\right)+c\sqrt{(a+c+7b)^3}\left((b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2\right)}=0$$ or $$(b-c)\sum_{sym}(19a^5+83a^4b+154a^3b^2-716a^3bc-2811a^2b^2c)+$$ $$+\tfrac{3(b-c)A\sqrt{(b+c+7a)^3}}{b\sqrt{(a+b+7c)^3}\left((b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2\right)+c\sqrt{(a+c+7b)^3}\left((b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2\right)}=0,$$ Where $$A=361(b+c)a^7+(2071b^2+9956bc+2071c^2)a^6+(5445b^3+42987b^2c+42987bc^2+5445c^3)a^5+$$ $$+(8507b^4+90734b^3c+146742b^2c^2+90734bc^3+8507c^4)a^4+$$ $$+(8507b^5+113647b^4c+397478b^3c^2+397478b^2c^3+113647bc^4+8507c^5)a^3+$$ $$+(5445b^6+87672b^5c+453747b^4c^2+867456b^3c^3+453747b^2c^4+87672bc^5+5445c^6)a^2+$$ $$+(2071b^7+39613b^6c+260655b^5c^2+691709b^4c^2+691709b^3c^4+260655b^2c^5+39613bc^6+2071c^7)a+$$ $$+361b^8+8246b^7c+67456b^6c^2+232490b^5c^3+314734b^4c^4+232490b^3c^5+67456b^2c^6+8246bc^7+361c^8.$$

If we'll prove that: $$\sum_{sym}(19a^5+83a^4b+154a^3b^2-716a^3bc-2811a^2b^2c)+$$ $$+\tfrac{3A\sqrt{(b+c+7a)^3}}{b\sqrt{(a+b+7c)^3}\left((b+c+7a)^2+(b+c+7a)(a+c+7b)+(a+c+7b)^2\right)+c\sqrt{(a+c+7b)^3}\left((b+c+7a)^2+(b+c+7a)(a+b+7c)+(a+b+7c)^2\right)}>0,$$ so it's enough to prove our inequality for equality case of two variables.

I hope it will help.