Ta có:
\(\dfrac{a^2}{a+2b^2}=a-\dfrac{2ab^2}{a+2b^2}=a-\dfrac{2ab^2}{a+b^2+b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a.b^2.b^2}}=a-\dfrac{2ab^2}{3\sqrt[3]{ab}.b}=a-\dfrac{2ab}{3\sqrt[3]{ab}}=a-\dfrac{2}{3}\sqrt[3]{a^2b^2}\ge a-\dfrac{2}{3}.\dfrac{a+b+ab}{3}=a-\dfrac{2}{9}\left(a+b+ab\right)\)
Tương tự: \(\dfrac{b^2}{b+2c^2}=b-\dfrac{2}{9}\left(b+c+bc\right)\)
\(\dfrac{c^2}{c+2a^2}\ge c-\dfrac{2}{9}\left(c+a+ca\right)\)
Cộng vế theo vế các BĐT vừa chứng minh, ta được:
\(VT\ge a+b+c-\dfrac{2}{9}.2\left(a+b+c\right)-\dfrac{2}{9}\left(ab+bc+ca\right)\)
\(VT\ge3-\dfrac{4}{9}.3-\dfrac{2}{9}.\dfrac{\left(a+b+c\right)^2}{3}=1\)
Vậy ta đpcm. Đẳng thức xảy ra khi \(a=b=c=1\)
#Chiến binh Alpha :))
