Bài toán:   Cho $a,b,c>0$. CMR:
$$\dfrac{a^{2}+b^{2}}{a+b}+\dfrac{b^{2}+c^{2}}{b+c}+\dfrac{c^{2}+a^{2}}{c+a}\leq 3\left(\dfrac{a^{2}+b^{2}+c^{2}}{a+b+c}\right)$$

Giải:
$$BDT\iff \dfrac{a^{2}+b^{2}}{a+b}+\dfrac{b^{2}+c^{2}}{b+c}+\dfrac{c^{2}+a^{2}}{c+a}\leq 3(\dfrac{a^{2}+b^{2}+c^{2}}{a+b+c}) \\ \iff (a+b+c)\left[\dfrac{a^{2}+b^{2}}{a+b}+\dfrac{b^{2}+c^{2}}{b+c}+\dfrac{c^{2}+a^{2}}{c+a}\right]\leq 3(a^{2}+b^{2}+c^{2}) \\ \iff \dfrac{c(a^{2}+b^{2})}{a+b}+\dfrac{a(b^{2}+c^{2})}{b+c}+\dfrac{b(c^{2}+a^{2})}{c+a}\leq a^{2}+b^{2}+c^{2} \\ \iff {a^2} + {b^2} + {c^2} - \dfrac{{c\left( {{a^2} + {b^2}} \right)}}{{a + b}} - \dfrac{{a\left( {{b^2} + {c^2}} \right)}}{{b + c}} - \dfrac{{b\left( {{c^2} + {a^2}} \right)}}{{a + c}} \ge 0  \\ \iff \dfrac{{ab{{\left( {a - b} \right)}^2}}}{{\left( {a + c} \right)\left( {b + c} \right)}} + \dfrac{{bc{{\left( {b - c} \right)}^2}}}{{\left( {b + a} \right)\left( {c + a} \right)}} + \dfrac{{ca{{\left( {c - a} \right)}^2}}}{{\left( {c + b} \right)\left( {a + b} \right)}} \ge 0$$
Vì $a,b,c>0$ nên ta được điều phải chứng minh
Dấu = xảy ra $ \Leftrightarrow a = b = c.\blacksquare$