Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2} \text{ when } abc=1$

Question

So I have a possible proof but I'm not certain it's right:

$\text{As } \frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}\text{, we obtain } \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{a^3(b+c) + b^3(a+c) + c^3(a+b)}$

$\begin{align} a^3(b+c) + b^3(a+c) + c^3(a+b) & = a^2(\frac{1}{b} + \frac{1}{c}) + b^2(\frac{1}{a} + \frac{1}{c}) c^2(\frac{1}{a} + \frac{1}{b}) \\ & \geq \frac{4a^2}{b+c} + \frac{4b^2}{a+c} + \frac{4c^2}{a+b} \text{ by the same inequality as above} \\ & \geq 4\frac{(a+b+c)^2}{2(a+b+c)} \text{ (same reason)} \\ & \geq 2(a+b+c) \end{align}$

Now, we have:

$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{2(a+b+c)}$

I'm not sure if this is a logical progression to make though. Because if $a \geq b$ then $\frac{1}{a} \leq \frac{1}{b}$, so is this not true? I feel like it must be as I can get the desired result, but I'm confused about it.

However, if this is true it is then sufficient to prove that $a+b+c \geq 3$.

$a+b+c = a+b + \frac{1}{ab} \geq 2\sqrt{ab} + \frac{1}{ab}$

It can be proved with calculus that $\forall x\ge 0,f(x) = 2\sqrt{x} + \frac{1}{x} \geq 3$. Thus the inequality is true, and the proof is finished.

Can someone please explain either why my proof is true (namely what I am confused about) or why this false proof yields the right result?

Answer 1

I think your first step gives a wrong inequality.

For the proof we can make the following thinks.

By Holder and AM-GM we obtain: $$\sum_{cyc}\frac{1}{a^3(b+c)}=\sum_{cyc}\frac{b^3c^3}{b+c}\geq\frac{(ab+ac+bc)^3}{3\sum\limits_{cyc}(b+c)}=$$ $$=\frac{(ab+ac+bc)(ab+ac+bc)^2}{6(a+b+c)}\geq\frac{3(ab+ac+bc)^2}{6(a+b+c)}=\frac{(ab+ac+bc)^2}{2abc(a+b+c)}\geq\frac{3}{2},$$ where the last inequality it's just $\sum\limits_{cyc}c^2(a-b)^2\geq0$.

Done!

Answer 2

I don't think generally $\frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$($a_i=10$, $x_i=1$ is counterexample), it should have been$$\frac{a_1^2}{x_1} + \frac{a_2^2}{x_2} + \frac{a_3^2}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$$by Cauchy-Schwartz inequality.

And if $x<y$ then $1/x>1/y$, therefore you should calculate the upper bound of $a^3(b+c) + b^3(a+c) + c^3(a+b)$, not the lower bound.

Also, it is true that $a+b+c \ge 3$ and you can prove it much simply with AM-GM inequality: $a+b+c \ge 3\sqrt[3]{abc}=3$. However, it only proves that $\frac{9}{2(a+b+c)}\le \frac{3}{2}$, which is not what you want.

Answer 3

We have $$\sum \frac{1}{a^3(b+c)} = \sum \frac{1}{a^2(ab+ac)} = \sum\frac{(bc)^2}{ab+ca}.$$ Replacing $ab=z,bc=x,ca=y$, we need to show $$\sum\frac{x^2}{y+z}\ge\frac32,\tag1$$ for $xyz=1$.

Now (1) is trivial from C-S: $$2(x+y+z) \sum\frac{x^2}{y+z}=\sum (y+z) sum\frac{x^2}{y+z}\ge (x+y+z)^2$$ and $$x+y+z \ge 3\sqrt[3]{xyz}=3.$$