# Proving inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4$

I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly.

Exercise 1.3.4(a) Let $$a,b,c$$ be positive real numbers. Prove that $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.$$

(b) For real numbers $$a,b,c \gt0$$ and $$n \leq3$$ prove that: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

• Please avoid using $$__$$ in the title. Commented Aug 31, 2012 at 5:19

Write $$\frac ab+\frac ab+\frac bc\geq \frac{3a}{\sqrt[3]{abc}}$$ by AM-GM.

You get $$\operatorname{LHS} \geq \frac{a+b+c}{\sqrt[3]{abc}}+n\left(\frac{\sqrt[3]{abc}}{a+b+c}\right).$$

Set $$z:=\frac{a+b+c}{\sqrt[3]{abc}}$$ and then notice that for $n\leq 3$, $$z+\frac{3n}{z}\geq 3+n.$$ Indeed the minimum is reached for $z=\sqrt{3n}\leq 3$; since $z\geq 3$, the minimum is reached in fact for $z=3$.

• How did you prove that $z+3n/z \ge 3+n$ for $n\le 3?$ Commented Oct 12, 2021 at 6:39

By C-S $$\sum_{cyc}\frac{a}{b}=\sum_{cyc}\frac{a^2}{ab}\geq\frac{(a+b+c)^2}{ab+ac+bc}.$$ Thus, it's enough to prove that $$\frac{(a+b+c)^2}{ab+ac+bc}+\frac{3\sqrt[3]{abc}}{a+b+c}\geq4.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality it's $f(v^2)\geq0,$ where $f$ decreases, which says that it's enough to prove the last inequality for a maximal value of $v^2$, which happens for equality case of two variables.

Since the last inequality is homogeneous, we can assume $b=c=1$. Also, let $a=x^3$.

Id est, we need to prove that $$\frac{(x^3+2)^2}{2x^3+1}+\frac{3x}{x^3+2}\geq4$$ or $$(x-1)^2(x^6+2x^5+3x^4+2x^3+x^2+6x+3)\geq0.$$ Done!

The following inequality is also true.

Let $a$, $b$ and $c$ be positives. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{24\sqrt[3]{abc}}{a+b+c}\geq11.$$

• what is C-S? Can you explain Commented Aug 2, 2018 at 15:28
• It's Cauchy-Schwarz inequality: For all reals $a_i$ and positives $b_i$ we have: $\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+...+\frac{a_n^2}{b_n}\geq\frac{(a_1+a_2+...+a_n)^2}{b_1+b_2+...+b_n}.$ Commented Aug 2, 2018 at 16:02
• oh thanks, I didn't know that. Commented Aug 2, 2018 at 16:20

Following above motivations and applying AM-GM three times: \begin{align} &\frac13\left(\frac ab+\frac ab+\frac bc\right)+\frac13\left(\frac bc+\frac bc+\frac ca\right)+\frac13\left(\frac ca+\frac ca+\frac ab\right)+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &\ge \frac{a}{\sqrt[3]{abc}}+\frac{b}{\sqrt[3]{abc}}+\frac{c}{\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &=\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{a+b+c}{3\sqrt[3]{abc}}+\frac{3\sqrt[3]{abc}}{a+b+c}\\ &\ge4\left(\left(\frac{a+b+c}{3\sqrt[3]{abc}}\right)^2\right)^\frac14\\ &\ge4. \end{align}