Inequality involving AM-GM but its wierd [duplicate]

Question

Let a, b, c be positive real numbers. Prove that

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

Ohk now i know using AM-GM that $ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq 3 \cdot \sqrt[3] {\frac{a}{b} \cdot \frac{b}{c} \cdot \frac{c}{a}} = 3 \cdot 1=3 $

Now if i could have shown that the other term $\geq 1$. I would be done. But the problem is that (again using AM-GM)

$\sqrt[3]{abc} \leq \frac{a+b+c}{3} \Rightarrow 3 \cdot \sqrt[3]{abc} \leq a+b+c \Rightarrow \frac{3 \cdot \sqrt[3]{abc}}{a+b+c} \leq 1$

So if first part is $\geq 3$ and second part is $\leq 1$, How will i show that it is greater than $4$? Is my approach correct? Or is there something wrong with the question?

Thanks.

(Source: https://web.williams.edu/Mathematics/sjmiller/public_html/161/articles/Riasat_BasicsOlympiadInequalities.pdf ,pg-$13$ exercise $1.3.4.a$ )

Answer 1

From $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$ and $ \sqrt[3]{abc} \geqslant \frac{3abc}{ab+bc+ca},$ we need to prove $$\frac{(a+b+c)^2}{ab+bc+ca} + \frac{9abc}{(ab+bc+ca)(a+b+c)} \geqslant 4,$$ equivalent to $$ (a+b+c)^2 + \frac{9abc}{a+b+c} \geqslant 4(ab+bc+ca),$$ or $$a^2+b^2+c^2 + \frac{9abc}{a+b+c} \geqslant 2(ab+bc+ca).$$ Which is Schur is inequality.

Answer 2

Your steps are right, but they are don't give a proof.

You proved that $$A=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3$$ and $$B=\frac{3\sqrt[3]{abc}}{a+b+c}\leq1.$$ We need prove that $$A+B\geq4,$$ which is impossible by your work.

For example, for $A=3.1$ and $B=0.8$ we obtain $$A+B\geq4$$ is wrong.