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: $$\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.$$

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Please avoid using $$__$$ in the title. –  Sasha Aug 31 '12 at 5:19

First, I will prove that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$, and then I will prove that $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}$.

1) $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3$

Let's call $\varphi(x_1,x_2,x_3)=\frac{x_1}{x_2}+\frac{x_2}{x_3}+\frac{x_3}{x_1}$. Le'ts prove that $\varphi$ is minimized when $x_1 = x_2 = x_3$: $\frac{\partial\varphi}{\partial x_i} = \frac{1}{x_{i+1}} - \frac{x_{i-1}}{x_i^2} = \frac{x_i^2 - x_{i-1}x_{i+1}}{x_i^2 x_{i+1}}$ (with $x_4 = x_1$ and $x_0 = x_3$).

So $\Delta\varphi = 0$ is equivalent to : $$x_1^2 = x_2 x_3$$ $$x_2^2 = x_1 x_3$$ $$x_3^2 = x_1 x_2$$

By dividing one equation by another, we get that $x_1 = x_2 = x_3$. We need to prove that $\varphi$ is convex, to show that this function is minimized when its gradient is $0$. For this, let's compute its Hessian matrix $M$:

$$M = (M_{ij}) = (\frac{\partial^2\varphi}{\partial x_i\partial x_j})$$ We have: $$M_{ii} = \frac{2 x_{i-1}}{x_i^3}$$ $$M_{i(i+1)} = -\frac{1}{x_{i+1}^2}$$ $$M_{i(i-1)} = -\frac{1}{x_i^2}$$

Which gives us:

$$M = \left( \begin{array}{ccc} \frac{2 x_3}{x_1^3} && -\frac{1}{x_2^2} && -\frac{1}{x_1^2}\\ -\frac{1}{x_2^2} && \frac{2 x_1}{x_2^3} && -\frac{1}{x_3^2}\\ -\frac{1}{x_1^2} && -\frac{1}{x_3^2}&& \frac{2 x_2}{x_3^3} \end{array} \right)$$

Calculating its determinant and verifying it is positive will show that $\varphi$ is convex, which achives to prove inequality 1)

2) $\displaystyle\frac{3 \sqrt[3]{abc}}{a+b+c}\geq 1$ This is equivalent to $\sqrt[3]{abc}\geq\frac{a+b+c}{3}$, or $\ln(\frac{a+b+c}{3})\geq \frac{1}{3}(\ln a +\ln b+\ln c)$, which is true because the function $\ln$ is concave.

By combining those two inequalities, we instantly solve the two questions.

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in part 2 your inequality is in the wrong direction. By AM-GM arithmetic average is larger than geometric average. –  Maesumi Sep 12 '12 at 0:18
@Maesumi that's right, I made a mistake after taking the log, corrected now. Thanks for pointing it out! –  S4M Sep 12 '12 at 7:54
@S4M By AM-GM, $\sqrt[3]{abc}\leq\frac{a+b+c}{3}$. This is a well-known inequality. –  Arthur Sep 12 '12 at 7:59
-1 Part 2 is wrong, as pointed out by Arthur. @Iuli Please unaccept this. –  Calvin Lin Nov 11 '13 at 14:48

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 $z=3$.

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