Observe that a b and c are cot of angles of triangle. Then the problem reduces to a very famous problem

Let $ x= \sum \frac{1}{a^2+1} $ After using the condition we get: $x = 2 + \frac{2abc}{(a+b)(b+c)(c+a)}$ and thus by AM-GM and $a,b,c \geq 0$ then $\frac{9}{4} \geq x \geq 2$ But the expression in the problem is just $x(3-x)$ which is decreasing in the given intervall so max and min are at the endpoints. Result: $ 2 \geq x(3-x) \geq \frac{27}{16}$

XbenX wrote: 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 $ab+bc+ca=1$. Let $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1.$ Then $$\frac{27}{16}\leq\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\leq 2.$$Solution of Zhangyanzong:

Attachments:

$x,y,z>0,xy+yz+zx=1$,prove that \[ {\frac {27}{16}}\geq \left( \left( yz+1 \right) ^{-1}+ \left( xz+1 \right) ^{-1}+ \left( xy+1 \right) ^{-1} \right) \left( {\frac {yz}{yz+1}}+{\frac {xz}{xz+1 }}+{\frac {xy}{xy+1}} \right) \geq \frac{5}{4}+{\frac {21}{16}}\,\sqrt {3}xyz\]

XbenX wrote: 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 $ab+bc+ca=1$. We can also solve this problem using homogenization and $uvw$, but it is rather ugly.

answer:$\frac{27}{16}$ and $2$

Let $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that $$\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\right)\geq \frac{27}{28}$$$$\left( \frac{1}{a^2+2}+\frac{1}{b^2+2}+\frac{1}{c^2+2}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq \frac{27}{28}$$

Let $a,b>0 $ and $ ab=1.$ Prove that: $$ \left( \frac{1}{a+k}+\frac{1}{b+k}\right)\left( \frac{a}{a+k}+\frac{b}{b+k}\right)\geq\frac{4}{(k+1)^2}$$Where $k>0.$ $$ \left( \frac{1}{a+k}+\frac{1}{b+k}\right)\left( \frac{a}{b+k}+\frac{b}{a+k}\right)\geq\frac{4}{(k+1)^2}$$Where $k\in N^+. $

xzlbq wrote: $x,y,z>0,xy+yz+zx=1$,prove that \[ {\frac {27}{16}}\geq \left( \left( yz+1 \right) ^{-1}+ \left( xz+1 \right) ^{-1}+ \left( xy+1 \right) ^{-1} \right) \left( {\frac {yz}{yz+1}}+{\frac {xz}{xz+1 }}+{\frac {xy}{xy+1}} \right) \geq \frac{5}{4}+{\frac {21}{16}}\,\sqrt {3}xyz\] Smart one. And also difficult

Let $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that $$\left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+3}+\frac{b^2}{b^2+3}+\frac{c^2}{c^2+3}\right)\geq \frac{27}{40}$$

sqing wrote: Let $a,b>0 $ and $ ab=1.$ Prove that: $$ \left( \frac{1}{a+k}+\frac{1}{b+k}\right)\left( \frac{a}{a+k}+\frac{b}{b+k}\right)\geq\frac{4}{(k+1)^2}$$Where $k>0.$

Attachments:

Let $S = \frac{1}{a^2+1}+\frac{1}{b^2+1} +\frac{1}{c^2+1}$. It is easy to see that: $$ \left( \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)=3S-S^2 $$The key claim is the following. Claim: $\frac{9}{4} \ge S \ge 2 $ Proof: Note that $S$ can be equivalently transformed into the following expression: \begin{align*} \frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1} = \\ =\frac{1}{a^2+ab+bc+ca}+\frac{1}{b^2+ab+bc+ca}+\frac{1}{c^2+ab+bc+ca} = \\ =\frac{1}{(a+b)(a+c)}+\frac{1}{(b+c)(b+a)} +\frac{1}{(c+a)(c+b)} = \\ =\frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \end{align*} We begin with the lower bound. It is sufficient to show that: \begin{align*} S = \frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \ge 2 \\ 2(a+b+c) \ge 2(a+b)(b+c)(c+a) = 2((a+b+c)(ab+bc+ca)-abc) \\ 2(a+b+c) \ge 2(a+b+c) - 2abc \\ 2abc \ge 0 \end{align*}Last inequality is obviously true as $a;b;c$ are nonnegative reals. Therefore $S \ge 2$, which can be achieved by taking $a=0; b=1; c=1$. Now we focus on the upper bound. We proceed in similar manner. \begin{align*} \frac{9}{4} \ge \frac{2(a+b+c)}{(a+b)(b+c)(c+a)} \\ 9(a+b)(b+c)(c+a) \ge 8(a+b+c) \\ 9((a+b+c)(ab+bc+ca)-abc) \ge 8(a+b+c) \\ 9(a+b+c) - 9abc \ge 8(a+b+c) \\ a+b+c \ge 9abc \end{align*}Thus it sufficient to show that last inequality holds. It is easy to see that from AM-GM we have: $$ 1=ab+bc+ca \ge 3\sqrt[3]{a^2b^2c^2} \implies \frac{1}{3} \ge \sqrt[3]{a^2b^2c^2}$$On the other hand $a+b+c \ge 3\sqrt[3]{abc}$. It will be enough to show that $3\sqrt[3]{abc} \ge 9abc \implies \frac{1}{3} \ge \sqrt[3]{a^2b^2c^2}$, which holds from before established results. The equality can be achieved by taking $a =b=c=\frac{1}{\sqrt{3}}$. Consequently, the claim is proved. So our problem is reduced to find the minimum and maximum of $f(S) = 3S - S^2$ with $2.25 \ge S \ge 2$. It is easy to see that $f(S)$ has only one stationary point at $S=1.5$. Therefore the maximum and minimum of the function in this range will be at the end points. Note that: $$ f(2) = 3 \cdot 2 - 2^2 = 2 \quad \text{and} \quad f(2.25) = 2 \cdot 2.25 - 2.25^2 = \frac{27}{16} $$Therefore the maximum of our expression is $2$ and the minimum is $\frac{27}{16}$ .

Let $a,b$ and $c$ are non-negative real numbers satisfying $ab+bc+ca=1.$ Prove that $$\left( \frac{1}{a^2+\frac{1}{6}}+\frac{1}{b^2+\frac{1}{6}}+\frac{1}{c^2+\frac{1}{6}}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq \frac{9}{2}$$$$\left( \frac{1}{a^2+\frac{1}{4}}+\frac{1}{b^2+\frac{1}{4}}+\frac{1}{c^2+\frac{1}{4}}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq \frac{27}{7}$$$$\left( \frac{1}{a^2+\frac{1}{3}}+\frac{1}{b^2+\frac{1}{3}}+\frac{1}{c^2+\frac{1}{3}}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq \frac{27}{8}$$$$\left( \frac{1}{a^2+\frac{1}{2}}+\frac{1}{b^2+\frac{1}{2}}+\frac{1}{c^2+\frac{1}{2}}\right)\left( \frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\geq \frac{27}{10}$$