How to prove that $b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c)$

Question

I am stuck with the following problem.

If $a>0$, $b>0$, $c >0$ and not all equal then prove that: $$b^2c^2+c^2a^2+a^2b^2 \gt abc(a+b+c).$$

Additional info:I'm looking for solutions using AM-GM .

I don't know how to progress . I will be grateful if someone explains . Thanks in advance ..

Answer 1

HINT: use that $$x^2+y^2+z^2\geq xy+yz+zx$$ after my hint above we have $$a^2b^2+b^2c^2+c^2a^2\geq a^2bc+ab^2c+abc^2$$

Answer 2

$b^2c^2+c^2a^2\ge 2c(abc)$ etc.

Answer 3

\begin{eqnarray*} \frac{a^2b^2+a^2b^2+a^2c^2+a^2c^2}{4}> a^2bc \end{eqnarray*} & sum cyclic perms.

Answer 4

By AM-GM $$\sum_{cyc}a^2b^2=\frac{1}{2}\sum_{cyc}2a^2b^2=\frac{1}{2}\sum_{cyc}(a^2b^2+a^2c^2)\geq$$ $$\geq\frac{1}{2}\sum_{cyc}2\sqrt{a^2b^2\cdot a^2c^2}=\frac{1}{2}\sum_{cyc}2a^2bc=\sum_{cyc}a^2bc.$$

Answer 5

From C-S: $$\left(\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{a^2}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge \left(a+b+c\right)^2 \Rightarrow $$ $$\left(a^2b^2+b^2c^2+c^2a^2\right)^2\ge a^2b^2c^2(a+b+c)^2.$$ equality occurs when $a=b=c.$

Taking square root and $a\ne b\ne c$ will do.

Answer 6

Hint. By $AM = \frac{x_1 + \dots + x_n}{n}\leq\sqrt[m]{\tfrac{x_1^m +\dots+ x_n^m}{n}} = PM$ taking $x_1=ab; x_2=ac, x_3=bc, n=3, m=2$ after squaring you obtain necessary inequality