Bất đẳng thức $12$

Bài Toán: Chứng minh rằng với mọi $a;b;c>0$ ta có $$\frac{a(b+c)}{(b+c)^2+a^2}+\frac{b(a+c)}{(a+c)^2+b^2}+\frac{c(a+b)}{(a+b)^2+c^2}\leq \frac{6}{5}$$

Lời giải:$\sum \dfrac{a(b+c)}{a^{2}+(b+c)^{2}}=\sum \dfrac{a(b+c)}{a^{2}+\dfrac{1}{4}(b+c)^{2}+\dfrac{3}{4}(b+c)^{2}}\leq \sum \dfrac{a(b+c)}{a(b+c)+\frac{3}{4}(b+c)^{2}}$

Ta có: $1-\dfrac{a(b+c)}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$

$=\dfrac{3}{4}\cdot \dfrac{(b+c)^{2}}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$

$\Rightarrow 3-\sum \dfrac{a(b+c)}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$

$=\dfrac{3}{4}\sum \dfrac{(b+c)^{2}}{a(b+c)+\dfrac{3}{4}(b+c)^{2}}$

$\geq \dfrac{3}{4}\cdot \dfrac{4(a+b+c)^{2}}{\dfrac{3}{2}(a^{2}+b^{2}+c^{2})+\dfrac{7}{2}(ab+bc+ca)}$

$= 6\cdot \dfrac{(a+b+c)^{2}}{3(a+b+c)^{2}+ab+bc+ca}\geq 6\cdot \dfrac{(a+b+c)^{2}}{3(a+b+c)^{2}+\dfrac{(a+b+c)^{2}}{3}}$

$= \dfrac{9}{5}\Rightarrow \sum \dfrac{a(b+c)}{a(b+c)+\frac{3}{4}(b+c)^{2}}\leq \dfrac{6}{5}$