Question

Cho a,b,c là các số thực dương.Cmr

\(\sum\dfrac{a^3}{b^2-bc+c^2}\ge\dfrac{3\left(ab+bc+ca\right)}{a+b+c}\)

Answer 1

Từ \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow a+b+c\ge\dfrac{3\left(ab+bc+ca\right)}{a+b+c}\). Tức cần chứng minh

\(\dfrac{a^3}{b^2-bc+c^2}+\dfrac{b^3}{c^2-ac+a^2}+\dfrac{c^3}{a^2-ab+b^2}\ge a+b+c\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a^4}{ab^2-abc+ac^2}+\dfrac{b^4}{bc^2-abc+a^2b}+\dfrac{c^4}{a^2c-abc+b^2c}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{a^2b+a^2b+b^2c+bc^2+c^2a+ca^2-3abc}\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a+b+c\right)\left(a^2b+a^2b+b^2c+bc^2+c^2a+ca^2-3abc\right)\)

\(\Leftrightarrow a^4+b^4+c^4+abc\left(a+b+c\right)\ge ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)\)

Đúng theo Schur bậc 4