# Let $a,b,c > 1$, Show that $abc + \frac 1 {a}+\frac 1 {b} + \frac 1 {c} > a+ b+ c +\frac 1 {abc}$

I'm confused how to prove this inequality, I tried by multiplying both sides by abc to get ride of fraction which got me

$abc² + bc+ac+ab -(a+c+b)abc-1>0<=> abc²+bc+ac+ab-ab²-bc²-ca²-1 > 0$

I still can't see any way to to transform this inequality.

When i looked at the book solution i found that they come up with formula without explaining how they got there. here's book solution of the inequality

$(a-\frac 1 {b})(b-\frac 1 {c})(c-\frac 1 {a}) > 0$

What do i need to know so i can make this kind of transformation to the inequality.

which got me $$\;\ldots \;\; abc²+bc+ac+ab-ab²-bc²-ca²-1 > 0$$

You made a couple of mistakes in the intermediate calculations, which is fairly obvious since the expression you got in the end is not symmetric in $$\,a,b,c\,$$, as it should be.

What you would actually get, instead, is:

$$a^2b^2c^2 + ab+bc+ca - a^2bc-ab^2c-abc^2-1 \gt 0 \;\;\iff\;\; (a b - 1) (a c - 1) (b c - 1) \gt 0$$

[ EDIT ]   The factorization is easier to "see" by defining $$\,u=bc, v=ac, w=ab\,$$, then:

\begin{align} a^2b^2c^2 + ab+bc+ca - a^2bc-ab^2c-abc^2-1 &= uvw - uv-vw-wu+u+v+w -1 \\ &= (u-1)(v-1)(w-1) \end{align}

• ops didn't see that thanks! – Dota Apr 4 '18 at 2:56
• I still coudn't come up with your formula by myself ? can you go step by step how you got it if thats possible? – Dota Apr 4 '18 at 3:05
• @Dota Edited into the answer. Other than that, the expression can also be factored the hard way, by considering it as a quadratic in $a$ (for example) and finding its roots using the quadratic formula. – dxiv Apr 4 '18 at 3:10