(Crux) Let: $a,b,c>0$ such that $a+b+c=3$. Prove that: $$\frac{1}{9-ab}+\frac{1}{9-bc}+\frac{1}{9-ca} \leq \frac{3}{8}.$$
Solution:
Put: $x=bc,y=ca,z=ab$. So the inequality can be rewritten as:
$$\frac{1-x}{9-x}+\frac{1-y}{9-y}+\frac{1-z}{9-z} \geq 0$$
Or:
$$\frac{(1-x)(6+x)}{(9-x)(6+x)}+\frac{(1-y)(6+y)}{(9-y)(6+y)}+\frac{(1-z)(6+z)}{(9-z)(6+z)} \geq 0$$
WLOG, suppose: $$a \geq b \geq c \Rightarrow z \geq y\geq x$$
So by Chebyshev:
$$LHS \geq \left(\sum_{cyc}(1-x)(6+x)\right)\left(\sum_{cyc}\frac{1}{(9-x)(6+x)}\right)$$
So we just have to prove:
$$\sum_{cyc}(1-x)(6+x) \geq 0 \Leftrightarrow 5(ab+bc+ca) + (ab+bc+ca)^2 \leq 18+6abc$$
which is true by Schur.
My question is how can they know when to add $$(6+x),(6+y),(6+z)$$
in each fractions?
Thank you a lots!
Let $$\sum_{cyc}(1-x)(k+x)\geq0$$ is true.
Thus, $$\sum_{cyc}(k+(1-k)ab-a^2b^2)\geq0$$ or after homogenization
$$k(a+b+c)^4-3(k-1)(ab+ac+bc)(a+b+c)^2-27(a^2b^2+a^2c^2+b^2c^2)\geq0,$$ which after assuming $a=b=1$ and $c=0$ gives $$k\geq\frac{15}{4}.$$
Easy to show that for $k=\frac{15}{4}$ the inequality $$k(a+b+c)^4-3(k-1)(ab+ac+bc)(a+b+c)^2-27(a^2b^2+a^2c^2+b^2c^2)\geq0$$ is true.
If you want to use Chebyshov, so triples $$\left((1-x)(k+x),(1-y)(k+y),(1-z)(k+z)\right)$$ and $$\left(\frac{1}{(9-x)(k+x)},\frac{1}{(9-x)(k+x)},\frac{1}{(9-x)(k+x)}\right)$$ should be have the same ordering.
For $a\geq b\geq c$ we have $x\leq y\leq z$, for which we need $$(1-x)(k+x)\geq(1-y)(k+y)\geq(1-z)(k+z),$$ which is true for any $k\geq1$
because $$(1-x)(k+x)\geq(1-y)(k+y)$$ it's $$(y-x)(x+y+k-1)\geq0.$$
Also, we need $$\frac{1}{(9-x)(k+x)}\geq\frac{1}{(9-y)(k+y)}$$ or $$(9-x)(k+x)\leq(9-y)(k+y)$$ or $$(y-x)(x+y+k-9)\leq0$$ or $$c(3-c)\leq9-k$$,for which we need $$9-4(9-k)\leq0$$ and we obtain: $$\frac{15}{4}\leq k\leq\frac{27}{4}.$$
I hope now it's clear.