Problem 1952

Solution:

The required inequality is obviously equivalent to the inequality
$$\frac{a}{p+a^{2}}+\frac{b}{p+b^{2}}+\frac{c}{p+c^{2}} \leq \frac{27}{31}$$
where $a+b+c=1$ and $p=a b c+1$. We consider the function
$$f(x)=\frac{3(a+b+c)}{3 x+a^{2}+b^{2}+c^{2}}-\frac{a}{x+a^{2}}-\frac{b}{x+b^{2}}-\frac{c}{x+c^{2}}$$
We will prove that $f(x) \geq 0$ holds for all $x \geq a b+b c+c a$. Reducing the expression for $f(x)$ to a common denominator gives
$$f(x)=\frac{A x^{2}+B x+C}{\left(x+a^{2}\right)\left(x+b^{2}\right)\left(x+c^{2}\right)\left(3 x+a^{2}+b^{2}+c^{2}\right)}$$
where $A \geq 0 \geq C$. In fact, one easily obtains
$$\begin{gathered} A=2 a^{3}+2 b^{3}+2 c^{3}-a b(a+b)-a c(a+c)-b c(b+c) \geq 0 \\ C=-a b c\left[a\left(b^{3}+c^{3}\right)+b\left(c^{3}+a^{3}\right)+c\left(a^{3}+b^{3}\right)-2 a b c(a+b+c)\right] \leq 0 \end{gathered}$$
Note that it does not matter what $B$ is. Therefore, the polynomial $P(x)=A x^{2}+B x+C$ (if it is not identically 0) has two real roots, one positive (say $x=x_{0}$) and one negative, and it holds that $P(x) \leq 0$ for $0 \leq x \leq x_{0}$ and $P(x) \geq 0$ for $x \geq x_{0}$. We claim that $f(a b+b c+c a) \geq 0$. Indeed,
$$\begin{aligned} & f(a b+b c+c a) \\ & =\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}-\frac{a}{(a+b)(a+c)}-\frac{b}{(b+c)(b+a)}-\frac{c}{(c+a)(c+b)} \\ & =\frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)}-\frac{2(a b+b c+c a)}{(a+b)(b+c)(c+a)} \geq 0 \quad \text{since} \\ & \quad \frac{3(a+b+c)}{a^{2}+b^{2}+c^{2}+3(a b+b c+c a)} \geq \frac{9}{4(a+b+c)} \geq \frac{2(a b+b c+c a)}{(a+b)(b+c)(c+a)}, \end{aligned}$$
which is what we wanted. Therefore, $P(a b+b c+c a) \geq 0$, i.e. $x_{0} \leq a b+b c+c a$, from which it follows that $P(x) \geq 0$ and $f(x) \geq 0$ for all $x \geq a b+b c+c a$. In particular, $f(1+a b c) \geq 0$ since $1+a b c>1>a b+b c+c a$. Thus we have proved
$$\frac{a}{1+a b c+a^{2}}+\frac{b}{1+a b c+b^{2}}+\frac{c}{1+a b c+c^{2}} \leq \frac{3}{3+a^{2}+b^{2}+c^{2}+3 a b c}$$
It remains only to prove that $a^{2}+b^{2}+c^{2}+3 a b c \geq \frac{4}{9}$, which together with (1) will give the required inequality. Homogenization gives $9(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)+27 a b c \geq 4(a+b+c)^{3}$, which is equivalent to
$$5\left(a^{3}+b^{3}+c^{3}\right)+3 a b c \geq 3(a b(a+b)+a c(a+c)+b c(b+c))$$
The last inequality follows immediately from Schur's inequality. This finally completes the proof of the problem's statement.

Second solution. After homogenization, reducing to a common denominator and simplifying, the inequality reduces to a symmetric inequality which is proved directly by Muirhead's inequality:
$$\begin{aligned} & \frac{23}{2} T_{900}+122 T_{810}+260 T_{720}+282 T_{630}+193 T_{540}+\frac{547}{2} T_{711}+807 T_{620}+284 T_{531} \\ &+91 T_{522}-98 T_{441}-1669 T_{432}-557 T_{333} \geq 0 \end{aligned}$$
where $T_{ijk}$ is the symmetric sum $x^{i} y^{j} z^{k}+\cdots$.