Prove $\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} },$ if $a+b+c+abc=4.$

Question

Problem. Let $a,b,c\ge 0: ab+bc+ca>0$ and $a+b+c+abc=4.$ Prove that $$\color{black}{\sqrt{\frac{7a+2}{b+c}}+\sqrt{\frac{7b+2}{a+c}}+\sqrt{\frac{7c+2}{b+a}}\ge \frac{9\sqrt{2}}{2} .}$$ Source: KhuongTrang.

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A big trouble is equality holds at $(a,b,c)=(1,1,1);(0,2,2).$

I hope Jichen lemma helps but it gives wrong inequality $$\color{black}{\frac{7a+2}{b+c}+\frac{7b+2}{a+c}+\frac{7c+2}{b+a}\ge \frac{33}{2}}$$when $a=b=c=1.$

I think LM helps as this problem but it's complicated. Also, the Holder is not easy since we need a appropriate yield.

All idea and comment are welcome. Thanks for contribution.

Answer 1

\begin{align*} \sqrt{\frac{7 a + 2}{b + c}} + \sqrt{\frac{7 b + 2}{a + c}} + \sqrt{\frac{7 c + 2}{b + a}} &\ge \frac{9 \sqrt{2}}{2} \\ \text{To prove this inequality, we'll use the Cauchy-Schwarz inequality: } \\ |a| \cdot |7 a + 2| + |b| \cdot |7 b + 2| + |c| \cdot |7 c + 2| &\ge 54(a^2 + b^2 + c^2) + 24 \\ \text{Now, let's use the given conditions to simplify the left side of the inequality: } \\ a, b, c > 0 \quad \text{and} \quad a + b + c + a b c = 4 \quad \text{implies that:} \\ a, b, c < \frac{4}{a b c} \quad \text{so that:} \\ |a| = \sqrt{\frac{a}{b c}}, \quad |b| = \sqrt{\frac{b}{a c}}, \quad |c| = \sqrt{\frac{c}{a b}} \quad \text{and} \quad a^2 + b^2 + c^2 = \frac{16}{a b c} \quad \text{gives:} \\ |a| \cdot |7 a + 2| + |b| \cdot |7 b + 2| + |c| \cdot |7 c + 2| &= \frac{112}{a b c} \cdot \frac{7 a + 2}{b c} + \frac{112}{a b c} \cdot \frac{7 b + 2}{a c} + \frac{112}{a b c} \cdot \frac{7 c + 2}{b a} \\ &\ge \frac{54(16)}{a b c} + \frac{24}{a b c} = \frac{816}{a b c} \\ \text{Now, let's compare the left and right sides: } \\ \frac{816}{a b c} &\ge \frac{9 \sqrt{2}}{2} \\ \text{Square both sides: } \\ \frac{729}{2 a b c} &\ge \frac{81}{2} \\ \text{Multiply both sides by 2: } \\ 1458 &\ge 9 \cdot 91 \\ \text{Since the left side is greater than the right side, the inequality is true. } \end{align*}

I hope I was able to explain it. Thank you for such a beautiful problem to solve!

Answer 2

By Holder $$\left(\sum_{cyc}\sqrt{\frac{7a+2}{b+c}}\right)^2\sum_{cyc}\frac{(b+c)(7a+b+c)^3}{7a+2}\geq\left(\sum_{cyc}(7a+b+c)\right)^3=729(a+b+c)^3.$$ Thus, it's enough to prove that: $$18(a+b+c)^3\geq\sum_{cyc}\frac{(b+c)(7a+b+c)^3}{7a+2}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$18u^3\geq\sum_{cyc}\frac{(b+c)(2a+u)^3}{7a+2},$$ which after full expanding gives $f(v^2)\geq0,$ where $f$ is a concave function.

I got $f''(v^2)=1152-12096u<0$ (because from the condition by AM-GM we obtain: $u\geq1$).

Thus, since the condition does not depend on $v^2$, by $uvw$ it's enough to prove $f(v^2)\geq0$ for equality case of two variables.

Let $b=a$.

Thus, $c=\frac{4-2a}{a^2+1},$ where $0\leq a\leq2,$ which gives $$(2-a)(a-1)^2(7a^{12}+32a^{11}-69a^{10}+156a^9+762a^8+174a^7+1536a^6+2970a^5+1722a^4+9014a^3+3040a^2+6192a+1680)\geq0$$ and we are done!