Inequality $\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} + \frac{1}{abc} \ge min$ with $a+b+c = \frac{3}{2}$

Question

For arbitrary positive reals $a, b, c$, such that $a+b+c = \frac{3}{2}$, I want to find

$$\min(\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} + \frac{1}{abc})$$

$\frac{a^3}{bc} + \frac{b^3}{ac} \ge 2 \sqrt{\frac{(ab)^3}{abc^2}} = 2 \frac{ab}{c}$, and similarly for two other pairs, so

$\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} + \frac{1}{abc} \ge \frac{ab}{c} + \frac{ac}{b} + \frac{bc}{a} + \frac{1}{abc} \ge \frac{(ab)^2+(bc)^2+(ac)^2+ 1}{abc} \ge \frac{4\sqrt[4]{(abc)^4}}{abc} = 4$

But Lagrange multipliers giving the minimum of $\frac {19} {2}$ (and also the fact that the problem can't be so trivial) suggest that there is no use trying to find a tripple that results in $\frac{a^3}{bc} + \frac{b^3}{ac} + \frac{c^3}{ab} + \frac{1}{abc} = 4$, so another approach is needed.

It may boil down to simply using CS once and I just can't see it.

Answer 1

For $a=b=c=\frac{1}{2}$ we obtain a value $9.5$.

We'll prove that it's a minimal value.

Indeed, after homogenization we need to prove that: $$\frac{a^4+b^4+c^4}{abc\cdot\frac{a+b+c}{\frac{3}{2}}}+\frac{\left(\frac{a+b+c}{\frac{3}{2}}\right)^3}{abc}\geq\frac{19}{2}$$ or $$\frac{3(a^4+b^4+c^4)}{abc(a+b+c)}+\frac{16(a+b+c)^3}{27abc}\geq19,$$ which is true by Muirhead and AM-GM (in fact, the first one also easily follows from AM-GM): $$a^4+b^4+c^4\geq abc(a+b+c)$$ and $$(a+b+c)^3\geq27abc.$$

Answer 2

So we need to find maximal $k$ such that $$a^4+b^4+c^4 +1\geq kabc$$ for all positive $a,b,c$ with $a+b+c =3/2$. Puting $a=b=c =1/2$ we get $19/16\geq k/8$ so $k\leq 19/2$.

Let's check if $k_{\max} = 19/2$? Using Cauchy inequality twice and Am-Gm in the third line we get.

\begin{align}a^4+b^4+c^4 &\geq {1\over 3}(a^2+b^2+c^2)^2\\ &\geq {1\over 9}(a^2+b^2+c^2)(a+b+c)^2 \\ & \geq \sqrt[3]{a^2b^2c^2}\sqrt[3]{abc}(a+b+c)\\&=3abc/2\end{align}

So we are left to check if $$3abc/2 +1\geq 19abc/2 $$ is true i.e. if $8abc\leq 1$ is true. But this is true by AM-GM since $a+b+c =3/2$.

Answer 3

Let $a=e^p$, $b=e^q$ and $c=e^r$ for $p,q,r\in \mathbb{R}$. Then the expression becomes

$$f(p,q,r)=e^{3p-q-r}+e^{3q-p-r}+e^{3r-p-q}+e^{-p-q-r}$$

for $a+b+c=\frac{3}{2}$. Suppose (WLOG) $p\ge q\ge r$ and $p>r$. Then for small enough $\epsilon >0$, it is easy to show the inequality below by Karamata's Majorization Inequality:

$$f(p,q,r) \ge f(p-\epsilon,q,r+\epsilon)$$

Hence, the minimum value of the expression is held when $p=q=r$, $a=b=c$, and the expression equals to $\frac{19}{2}$ .