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

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 ($$x+y+z\geq 3\sqrt{xyz}=3$$)

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