Bất đẳng thức 24

Bài Toán:Cho $a,b,c$ không âm.Chứng minh rằng: $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq 1+\dfrac{3(a^3+b^3+c^3)}{2(a+b+c)(a^2+b^2+c^2)}$

Lời giải:Ta có: $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}-\dfrac{3}{2}$

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

$=\dfrac{1}{2}\sum (a-b)(\dfrac{1}{b+c}-\dfrac{1}{c+a})$

$=\dfrac{1}{2}\sum (a-b)\dfrac{a-b)}{(b+c)(c+a)}$

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

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

$=\dfrac{1}{2}(1-\dfrac{3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})})$

$=\dfrac{1}{2}\dfrac{(a+b+c)(a^{2}+b^{2}+c^{2})-3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}$

$=\dfrac{1}{2}\dfrac{\sum (a+b)(a-b)^{2}}{(a+b+c)(a^{2}+b^{2}+c^{2})}$

Ta có BĐT viết lại thành:

$\sum \dfrac{a}{b+c}-1-\dfrac{3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}$

$=\dfrac{\sum (a-b)^{2}}{2(b+c)(c+a)}+\dfrac{1}{2}\dfrac{\sum (a+b)(a-b)^{2}}{(a+b+c)(a^{2}+b^{2}+c^{2})}$

$=\sum (a-b)^{2}(\dfrac{1}{2(b+c)(c+a)}+\dfrac{a+b}{(a+b+c)(a^{2}+b^{2}+c^{2})})\geq 0$

$ \Rightarrow \sum \dfrac{a}{b+c}\geq 1+\dfrac{3(a^{3}+b^{3}+c^{3})}{(a+b+c)(a^{2}+b^{2}+c^{2})}$