Does Quadratic Forms Help?

Let $d=\min \{ a,b,c,d\}$ then $0 \le d \le 1$

And $p=ab+bc+ca$ then $3 \le p \le 6$

We have: $6=ab+bc+ca+d(a+b+c) \ge p+d.\sqrt{3p} \Rightarrow d \le \dfrac{6-p}{\sqrt{3p}}$ . So that: $\dfrac{1}{d^2+1} \ge \dfrac{3p}{p^2-9p+36}$

Now we define:

$x=bc, y=ca, z=ab$

So that: $x+y+z=p \ge 3$

And we have: $\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}=\dfrac{x}{x+yz}+\dfrac{y}{y+xz}+\dfrac{z}{z+xy}=A$

WLOG we can suppose that : $x \ge y \ge z$

So that: $\dfrac{y}{y+xz} \ge \dfrac{z}{z+xy}$ and $y+xz \le z+xy$

Use Chebyshev’s Inequality we have: $\dfrac{y}{y+xz}.(y+xz)+\frac{z}{z+xy}(z+xy) \le \dfrac{1}{2}.\Big(\dfrac{y}{y+xz} + \dfrac{z}{z+xy} \Big).(y+z+x(y+z))$ $\Rightarrow \dfrac{y}{y+xz} + \dfrac{z}{z+xy} \ge \dfrac{2}{x+1}$

And $\dfrac{x}{x+yz} \ge \dfrac{x}{x+\dfrac{(y+z)^2}{4}}=\dfrac{4x}{4x+(p-x)^2}$

So that: $A \ge \dfrac{4x}{4x+(p-x)^2}+\dfrac{2}{x+1}$

We will prove: $\dfrac{4x}{4x+(p-x)^2}+\dfrac{2}{x+1} \ge \dfrac{9}{p+3} \forall 0 \le x \le p, 3 \le p \le 6$

$\Leftrightarrow (3x-p)^2(2p-3-x) \ge 0$ (obvious true)

We need to prove: $\dfrac{9}{3+p}+\dfrac{3p}{p^2-9p+36} \ge 2$ with $3 \le p \le 6$

$\Leftrightarrow (6-p)(p-3)^2 \ge 0$ (obvious true)

Equality holds when $x=y=z=p/3, p=3$ or $p=6$ , that means $a=b=c=d=1$ or $a=b=c=\sqrt{2},d=0$