# Inequality involving AM-GM but its wierd [duplicate]

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

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

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

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)

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

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 $$1.3.4.a$$ )

• I am sorry @MartinR and Alexey but my tiny teen brain is too dumb to understand those proofs. Would be fantastic if i could get a slightly simplified explanation i would be highyl obliged :). Jun 3, 2020 at 18:09

From $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geqslant \frac{(a+b+c)^2}{ab+bc+ca},$$ and $$\sqrt{abc} \geqslant \frac{3abc}{ab+bc+ca},$$ 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.
You proved that $$A=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3$$ and $$B=\frac{3\sqrt{abc}}{a+b+c}\leq1.$$ 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.