# $\sum\limits_{cyc} \sqrt{\frac{a^3}{1+bc}} \geq 2$ for $a, b, c > 0$ which satisfies $abc=1$.

$$\displaystyle \sum_{cyc} \sqrt{\dfrac{a^3}{1+bc}} \geq 2$$ for $$a, b, c > 0$$ which satisfies $$abc=1$$.

My attempt: \begin{align} &\text{let } a=\frac{y}{x}, b=\frac{x}{z}, c=\frac{z}{y}. \\ &\text{Substituting for the original F.E.: }\displaystyle \sum_{cyc}\sqrt{\frac{(\frac{y}{x})^3}{1+\frac{x}{y}}} \geq 2 \text{ for }x, y, z\in \mathbb{R}^+. \\ &\therefore \text{ETS) }\displaystyle \sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}} \geq 2. \\ \ \\ & \text{Two ways to think: }\\ \ \\ & (1) \\ &\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \frac {x}{y}\Bigg) \geq \sum_{cyc} \frac{y^2}{x^2+xy} \\ &\text{ETS) }\displaystyle \sum_{cyc} \frac{y^2}{x^2+xy} \geq 2\Bigg( \sum_{cyc} \frac {x}{y} \Bigg). \ \\ &(2)\\ &\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \sqrt{\frac {x+y}{y}}\Bigg) \geq \sum_{cyc} \frac{y^3}{x^3} \\ &\text{ETS) }\displaystyle \sum_{cyc} \frac{y^3}{x^3} \geq 2\Bigg( \sum_{cyc} \sqrt{\frac{x+y}{y}} \Bigg). \end{align}

p.s. I think we can't use the AM-GM one, but I'll try.

• Have you found $(a,b,c)$ such that the sum is equal to $2$ ? Commented Oct 25, 2021 at 13:14

Using Titu's Lemma, $$\sum\sqrt{\frac{a^3}{1+bc}}=\sum \frac{a^2}{\sqrt{a+1}}\ge\frac{(a+b+c)^2}{\sum \sqrt{a+1}} \tag{1}$$ It remains to show, $$(a+b+c)^2\ge 2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})$$ Using A.M-G.M inequality, $$\sum \frac{(a+1)+1}{2}\ge \sum\sqrt{(a+1)\cdot 1}\implies a+b+c+6\ge2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})$$ It remains to show, $$(a+b+c)^2 \ge (a+b+c)+6\tag{2}$$ which is true, since $$a+b+c\ge 3\sqrt[3]{abc}=3.$$

Proof of $$(2)$$:$$(a+b+c)^2\ge 3(a+b+c)\ge (a+b+c)+6$$

Given below is a proof of a stronger result.

Claim: $$\sum \sqrt{\frac{a^3}{1+bc}}\ge \frac{3}{\sqrt{2}}$$ Proceeding from $$(1)$$, it remains to show, $$\sqrt{2}(a+b+c)^2\ge3\sum{\sqrt{a+1}}\implies 4(a+b+c)^2\ge 3\sum2\sqrt{2(a+1)}$$ Using A.M-G.M inequality, $$\frac{\sum(a+1)+2}{2}\ge \sum \sqrt{2(a+1)}\implies a+b+c+9\ge \sum2\sqrt{2(a+1)}$$ It remains to show, $$4(a+b+c)^2\ge 3(a+b+c)+27$$ which is true, since $$a+b+c\ge 3.$$

Also, equality can be achieved, unlike the original problem, at $$a=b=c=1$$.

• +1 retracing back the steps it seems the equality is impossible Commented Oct 25, 2021 at 13:24
• Yes, that was my reasoning as well. Is it obvious what the minima is ? If $a=b=c$, we can achieve $\frac{3}{\sqrt{2}}$. Commented Oct 25, 2021 at 13:27
• I don't know, even if it is it might be difficult to prove Commented Oct 25, 2021 at 13:29
• You can modify your approach accordingly to prove the strengthened result. Hint in white font: $\color{white}{\frac{ (a+1) + 2 } { 2} \geq \sqrt{ (a+1) \cdot 2 } }$. Ping me if you're stuck. Commented Oct 25, 2021 at 14:11
• Thanks for the solution!
– RDK
Commented Oct 26, 2021 at 8:54