Also given at the Romanian JBTST, 2005: http://www.mathlinks.ro/Forum/viewtopic.php?p=232780#p232780

$ab+cb+ca \leq \frac{3}{4}\Longleftrightarrow (ab+bc+ca)(a+b+c) \leq \frac{3(a+b+c)}{4}$ $\Longleftrightarrow \frac{3(a+b+c)}{4}\geq 1+abc$. To kill the problem, see that: i) $8abc \leq (a+b)(b+c)(c+a) = 1 \Longrightarrow abc \leq \frac{1}{8}$. ii)$2(a+b+c) = (a+b)+(b+c)+(c+a) \geq 3((a+b)(b+c)(c+a))^{\frac{1}{3}}= 3$

Let $x=a+b,y=b+c,z=c+a$ then $xyz=1$ Rewrite ineq in form $2xy+2yz+zx-x^{2}-y^{2}-z^{2}\leq 3$ Using mixing variable, ineq is proved

eagerkill wrote: Let $x=a+b,y=b+c,z=c+a$ then $xyz=1$ Rewrite ineq in form $2xy+2yz+zx-x^{2}-y^{2}-z^{2}\leq 3$ Using mixing variable, ineq is proved Oh.... it very interesting because it is a problem in VMO 2006, group B, too We can solve it by schur or assume $a,b\le 1$ or $a,b\ge 1$... It was posted long time ago by Darij .

My idea: $a+b=\frac{x^2}{yz}, a+c=\frac{y^2}{ax}, b+c={z^2}{xy}$, summing up we get $a+b+c=\frac{x^2}{2yz}+\frac{y^2}{2xz}+\frac{z^2}{2xy}$, and we easy get $c=\frac{y^3+z^3-x^3}{2xyz}$. The inequality can be written as: $\displaystyle\sum\limits_{\text{cyc}}x^6+3x^2y^2z^2\ge2(x^3y^3+x^3z^3+y^3z^3)$, but this is easy using AM-GM and Vornicu-Scur in the form $\displaystyle\sum\limits_{\text{cyc}}x^2(x^2-y^2)(x^2-z^2)\ge0$

A bit overkill Let $f(a,b,c)=\sum_{cyc}ab\text{ and }g(a,b,c)=\sum_{sym}a^2b+2abc-1$ furthermore define $L=f(a,b,c)+\lambda g(a,b,c)$ $$\frac{\partial L}{\partial a}=b+c+\lambda\left(b^2+c^2+2\sum_{cyc}ab\right)=0\Longrightarrow b+c+\lambda b^2+\lambda c^2=-2\lambda\sum_{cyc}ab, (1)$$$$\frac{\partial L}{\partial b}=c+a+\lambda\left(c^2+b^2+2\sum_{cyc}ab\right)=0\Longrightarrow c+a+\lambda c^2+\lambda a^2=-2\lambda\sum_{cyc}ab, (2)$$$$\frac{\partial L}{\partial c}=a+b+\lambda\left(a^2+b^2+2\sum_{cyc}ab\right)=0\Longrightarrow a+b+\lambda a^2+\lambda c^2=-2\lambda\sum_{cyc}ab, (3)$$ From $(1)\text{ and }(2)$ we obtain $\lambda(a-b)(a+b)=-(a-b)$ From $(2)\text{ and }(2)$ we obtain $\lambda(b-c)(b+c)=-(b-c)$ From $(1)\text{ and }(3)$ we obtain $\lambda(a-c)(a+c)=-(a-c)$ Furthermore $FTSOC$ assume $a\neq b\neq c$, thus we obtain $-\frac{1}{\lambda}=a+b=b+c=c+a\Longrightarrow a=b=c$ which is clearly a contradiction, thus $a=b=c$. Plugging this in $g(a,a,a)$ we obtain $8a^3=1\Longrightarrow a^3=\frac{1}{8}\Longrightarrow a=b=c=\frac{1}{2}$ therefore maximum is achieved in the assertion $f\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)=\frac{3}{4}$ $\blacksquare$.

better solution . $(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc=1$ $ab+bc+ca=\frac{1+abc}{a+b+c}$ $a^2b+a^2c+b^2a+c^2a+b^2c+c^2b \geq 6 abc $ Then $\frac{1}{8} \geq abc $ we need to prove $\frac{1+abc}{a+b+c} \le \frac{3}{4}$ $\frac{1+abc}{a+b+c} \le \frac{\frac{3}{8}}{a+b+c} \le \frac{1}{4}$ $2(a+b+c)=(a+b)+(b+c)+(c+a) \geq 3$ ($AM-GM$) so $a+b+c \geq \frac{3}{2}$ $ \frac{\frac{3}{8}}{a+b+c} \le \frac{1}{4}$ and that is indeed true.

Maybe we can use AM-GM somewhere Also @2above why 12 year bump