Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

Question

I can't make progress with proving this inequality.

I have tried opening the brackets and using $abc=1$ in order to obtain the following:

$$a^2+b^2+ \frac 1{a^2b^2}+18+9a^2b^2+\frac 9{a^2}+\frac9{b^2}\ge 16 \left (ab+\frac 1a+\frac 1b \right)$$

Answer 1

We begin by observing that if $a=b=c=1$, then the two expressions are equal to each other.

Expand everything. Use the equation $abc=1$ to get rid of compound terms. Gather terms with each variable separately. Use the preceding observation to factor each nominator.

$$\begin{align} & 9(1+a^2)(1+b^2)(1+c^2)-8(a+b+c)^2 \\ &=9\bigl(1+a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+a^2b^2c^2\bigr)-8\bigl(a^2+b^2+b^2+2ab+2bc+2ca\bigr) \\ &= 9\Bigl(1+a^2+b^2+c^2+{1\over c^2}+{1\over a^2}+{1\over b^2}+1\Bigr)-8\Bigl(a^2+b^2+c^2-{2\over c}-{2\over a}-{2\over b}\Bigr) \\ & ={a^4+6a^2-16a+9 \over a^2}+{b^4+6b^2-16b+9 \over b^2}+{c^4+6c^2-16c+9 \over c^2} \\ & ={(a-1)^2(a^2+2a+9) \over a^2}+{(b-1)^2(b^2+2b+9) \over b^2}+{(c-1)^2(c^2+2c+9) \over c^2}\end{align}$$ It is now obvious that this last expression is always $\geq 0$, and that it is $=0$ only in the case when $a=b=c=1$.

Answer 2

Since the right side of your inequality does not depend on substitution $a\rightarrow-a$,

it's enough to prove it for positive variables.

We'll prove that your inequality is true for all positives $a$, $b$ and $c$ such that $abc=1$.

Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, our inequality is equivalent to $$8\cdot9u^2w^2\leq18w^4+9(9u^2-6v^2)w^2+9(9v^4-6uw^3)$$ or

$f(v^2)\geq0$, where $$f(v^2)=9v^4-6v^2w^2-6uw^3+u^2w^2+2w^4$$ Since $f'(v^2)=18v^2-6w^2>0$, it remains to prove our inequality for a minimal value of $v^2$,

which happens for equality case of two variables.

Let $b=a$ and $c=\frac{1}{a^2}$.

We obtain $(a-1)^2(9a^6+18a^5+13a^4+8a^3+21a^2+2a+1)\geq0$.

Done!