Question

Cho a , b , c , d > 0 . Cmr

\(\Sigma\dfrac{a^3}{b+c+d}\ge\dfrac{a^2+b^2+c^2+d^2}{3}\)

Answer 1

\(A=\dfrac{a^3}{b+c+d}+\dfrac{b^3}{a+c+d}+\dfrac{c^3}{a+b+d}+\dfrac{d^3}{a+b+c}\)

\(=\dfrac{a^4}{ab+ac+ad}+\dfrac{b^4}{ab+bc+bd}+\dfrac{c^4}{ac+bc+cd}+\dfrac{d^4}{ad+bd+cd}\)

\(\ge\dfrac{\left(a^2+b^2+c^2+d^2\right)^2}{2\left(ab+ac+ad+bc+bd+cd\right)}\) (bđt Cauchy Shwarz dạng Engel)

Cần chứng minh \(\dfrac{a^2+b^2+c^2+d^2}{2\left(ab+ac+ad+bc+bd+cd\right)}\ge\dfrac{1}{3}\)

\(\Leftrightarrow3a^2+3b^2+3c^2+3d^2\ge2\left(ab+ac+ad+bc+bd+cd\right)\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(a-d\right)^2+\left(b-d\right)^2+\left(b-c\right)^2+\left(c-d\right)^2\ge0\) *đúng*

Vậy ta có đpcm.

Dấu "=" xảy ra khi a = b = c = d