## Problem T-1

Determine the smallest and the greatest possible values of the expression

$$
\left(\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}\right)\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right)
$$

provided $a, b$, and $c$ are non-negative real numbers satisfying $a b+b c+c a=1$.

(proposed by Walther Janous, Austria)

Answer. The smallest value is $\frac{27}{16}$ and the greatest is 2 .


Solution. Let us denote

$$
x=\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}, \quad y=\frac{2 a b c}{(a+b)(b+c)(c+a)}
$$

to simplify notation.

Denominators in $x$ can be manipulated using $a b+b c+c a=1$ as

$$
a^{2}+1=a^{2}+a b+b c+c a=(a+b)(a+c)
$$

and similarly for $b^{2}+1$ and $c^{2}+1$. This yields a relation between $x$ and $y$

$$
x=\frac{a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)}{(a+b)(b+c)(c+a)}=1-\frac{2 a b c}{(a+b)(b+c)(c+a)}=1-y
$$

Using

$$
\frac{1}{a^{2}+1}=1-\frac{a^{2}}{a^{2}+1}
$$

and similar relations for $b$ and $c$ we have

$$
\left(\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{b^{2}+1}\right)\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right)=(3-x) x
$$

Since $(3-x) x=(2+y)(1-y)=2-y-y^{2}$, we want to estimate $y$. Obviously $y \geq 0$ with equality e.g. for $(a, b, c)=(0,1,1)$. On the other hand $y \leq \frac{1}{4}$ as can be seen after multiplying three AM-GM inequalities

$$
a+b \geq 2 \sqrt{a b}, \quad b+c \geq 2 \sqrt{b c}, \quad c+a \geq 2 \sqrt{c a}
$$

The equality is reached for $a=b=c=\sqrt{\frac{1}{3}}$.

Finally we compute

$$
\frac{27}{16}=2-\frac{1}{4}-\frac{1}{16} \leq 2-y-y^{2} \leq 2
$$

Similar solution. Let us denote

$$
x=\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}
$$

Since

$$
\frac{a^{2}}{a^{2}+1}=1-\frac{1}{a^{2}+1}
$$

holds we have

$$
\left(\frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}+\frac{1}{c^{2}+1}\right)\left(\frac{a^{2}}{a^{2}+1}+\frac{b^{2}}{b^{2}+1}+\frac{c^{2}}{c^{2}+1}\right)=x(3-x)=\frac{9}{4}-\left(x-\frac{3}{2}\right)^{2}
$$

To obtain the extrema it is sufficient to find the bounds of $x$.

The sum $a+b+c$ is non-negative because $a, b, c \geq 0$. If $a+b+c=0$ then $a=b=c=0$ what is in contradiction with $a b+b c+c a=1$, so $a+b+c>0$. Using $a b+b c+c a=1$ we obtain

$$
\frac{1}{a^{2}+1}=\frac{1}{a^{2}+a b+b c+c a}=\frac{1}{(a+b)(a+c)}
$$

and similarly for $1 /\left(b^{2}+1\right)$ and $1 /\left(c^{2}+1\right)$. This yields

$$
x=\frac{2(a+b+c)}{(a+b)(b+c)(c+a)}=\frac{2(a+b+c)}{(a+b+c)(a b+b c+c a)-a b c}=\frac{2}{1-\frac{a b c}{a+b+c}}
$$

and the problem is reduced to find bounds of $a b c /(a+b+c)$. We have obviously

$$
0 \leq \frac{a b c}{a+b+c}
$$

The equality holds there if one of $a, b, c$ is equal 0 and the product of remaining two is 1 . On the other hand using AM-GM inequality we have

$$
a+b+c=(a+b+c)(a b+b c+c a) \geq 3(a b c)^{\frac{1}{3}} \cdot 3(a b c)^{\frac{2}{3}}=9 a b c
$$

with equality in the case $a=b=c=\frac{\sqrt{3}}{3}$. So

$$
\frac{a b c}{a+b+c} \leq \frac{1}{9}
$$

These inequalities give us the bounds for $x$

$$
2 \leq x=\frac{2}{1-\frac{a b c}{a+b+c}} \leq \frac{9}{4}
$$

This follows

$$
\frac{27}{16} \leq \frac{9}{4}-\left(x-\frac{3}{2}\right)^{2} \leq 2
$$

The lower bound arises if one of $a, b, c$ is zero and product of the others is 1 , the upper bound arises if $a=b=c=\frac{\sqrt{3}}{3}$.

