Oops forgot that $a+b$ might be negative.

The things inside the two brackets are the same......

youthdoo wrote: The things inside the two brackets are the same...... I don't think so.

Solution: Tintarn wrote: Real numbers $a,b,c$ satisfy $a+b \ne 0$, $b+c \ne 0$ and $c+a \ne 0$. Show that \[\left(\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a}\right) \cdot \left(\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a}\right) \ge 0.\] From T2 Lemma: $\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a} = \frac{a^2}{\frac{1}{c}(a + b)} + \frac{b^2}{\frac{1}{a}(b + c)} + \frac{c^2}{\frac{1}{b}(a + c)} \geq \frac {(a + b + c)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)}$ $\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a} = \frac{b^2}{\frac{1}{c}(a + b)} + \frac{c^2}{\frac{1}{a}(b + c)} + \frac{a^2}{\frac{1}{b}(a + c)} \geq \frac {(b + c + a)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)}$ So $$ (\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a}) * (\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a}) \geq (\frac {(a + b + c)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)})^2 \geq 0$$Equality holds when $$ \frac{a}{\frac{1}{c}(a + b)} = \frac{b}{\frac{1}{a}(b + c)} = \frac{c}{\frac{1}{b}(a + c)} $$$$ \frac{b}{\frac{1}{c}(a + b)} = \frac{c}{\frac{1}{a}(b + c)} = \frac{a}{\frac{1}{b}(a + c)} $$

$\frac {a^2c}{a+b} = c(a-b) + \frac {b^2c}{a+b}$ $\sum c(a-b)=0$, so we have $\sum \frac {a^2c}{a+b} = \sum \frac {b^2c}{a+b}$ Hence $\sum \frac {a^2c}{a+b} \cdot \sum \frac {b^2c}{a+b} = \big( \sum \frac{b^2c}{a+b}\big)^2 \ge 0$

Equality holds when $ ?$

Eagle116 wrote: Solution: Tintarn wrote: Real numbers $a,b,c$ satisfy $a+b \ne 0$, $b+c \ne 0$ and $c+a \ne 0$. Show that \[\left(\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a}\right) \cdot \left(\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a}\right) \ge 0.\] From T2 Lemma: $\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a} = \frac{a^2}{\frac{1}{c}(a + b)} + \frac{b^2}{\frac{1}{a}(b + c)} + \frac{c^2}{\frac{1}{b}(a + c)} \geq \frac {(a + b + c)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)}$ $\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a} = \frac{b^2}{\frac{1}{c}(a + b)} + \frac{c^2}{\frac{1}{a}(b + c)} + \frac{a^2}{\frac{1}{b}(a + c)} \geq \frac {(b + c + a)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)}$ So $$ (\frac{a^2c}{a+b}+\frac{b^2a}{b+c}+\frac{c^2b}{c+a}) * (\frac{b^2c}{a+b}+\frac{c^2a}{b+c}+\frac{a^2b}{c+a}) \geq (\frac {(a + b + c)^2}{\frac{1}{c}(a + b) + \frac{1}{a}(b + c) + \frac{1}{b}(a + c)})^2 \geq 0$$ You cannot apply Titu's lemma, since the denominators $\frac{a+b}{c}$ etc. need to be positive. In general, I doubt there is a solution using this lemma.

sqing wrote: Equality holds when $ ?$ For $\sum\limits_{cyc} \frac {a^2c}{a+b}=0.$