Proof verification: For $a$, $b$, $c$ positive with $abc=1$, show $\sum_{\text{cyc}}\frac{1}{a^3(b+c)}\geq \frac32$

Question

I would like to have my solution to IMO 95 A2 checked. All solutions I've found either used Cauchy-Schwarz, Chebyshevs inequality, the rearrangement inequality or Muirheads inequality. Me myself, I've used Jensens inequality, and I am a bit unsure if my solution holds. Here it is:

Let $a,b,c$ be positive real numbers with $abc=1$. Show that $$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)}\geq \frac32$$

First, substitute $\frac1a = x, \frac1b = y, \frac1c = z$, giving the new constraint $xyz=1$, and transforming the inequality into $$\frac{x^2}{z+y} + \frac{y^2}{x+z} + \frac{z^2}{x+y} \geq \frac32$$ Now, let $f(x)=\dfrac{x^2}{S-x}$ where $S=x+y+z$. We have that $f''(x)=\dfrac{2S^2}{(S-x)^2}$, and $S>0$ beacuse $x,y,z>0$, thus $f''(x)>0$ for all $x,y,z\in \mathbb{R}^+$. Thus we see that $f(x)$ is convex, so by Jensens inequality we have that $$\begin{alignat*}{2}\frac{x^2}{y+z} + \frac{y^2}{x+z} + \frac{z^2}{x+y} & =\frac{x^2}{S-x} + \frac{y^2}{S-y} + \frac{z^2}{S-z} \\ &\geq 3f\left(\frac{x+y+z}{3} \right) \\ &= 3\frac{\left(\frac{x+y+z}{3}\right)^2}{S-\frac{x+y+z}{3}} = \frac{1}{3}\frac{(x+y+z)^2}{\frac{2x+2y+2z}{3}} = \frac13 \frac{(x+y+z)^2}{\frac23 (x+y+z)} \\ &= \frac12 (x+y+z) \geq \frac32\end{alignat*}$$ The final inequality follows by AM-GM ($x+y+z\geq 3\sqrt[3]{xyz}=3$)

Answer 1

Your proof is nearly correct: $$f''(x)=\frac{2S^2}{(S-x)^3}$$ (the exponent is $3$, not $2$). This is still okay because $x,y,z>0$, but you need a bit more justification.

Answer 2

You make a little mistake i think: $$f''(x)=\frac{2x^2}{(S-x)^3}$$ (and the exponent is $3$, not $2$ in a numerator). Apart from that it is ok.

Answer 3

I think it's better to end your proof by C-S and AM-GM: $$\sum_{cyc}\frac{x^2}{y+z}\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(y+z)}=\frac{1}{2}(x+y+z)\geq\frac{3}{2}.$$