Question

chứng minh rằng : Với mọi số dương a, b, c, d ta có:

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

Answer 1

Xét BĐT phụ  \(\frac{a^3}{a^2+b^2}\ge\frac{2a-b}{2}\)\(\Leftrightarrow b\left(a-b\right)^2\ge0\)

Tương tự ta có:

\(\frac{b^3}{b^2+c^2}\ge\frac{2b-c}{2};\frac{c^3}{c^2+d^2}\ge\frac{2c-d}{2};\frac{d^3}{d^2+a^2}\ge\frac{2d-a}{2}\)

Cộng lại theo vế ta có:

\(VT\ge\frac{2a-b}{2}+\frac{2b-c}{2}+\frac{2c-d}{2}+\frac{2d-a}{2}\)

\(=\frac{2a-b+2b-c+2c-d+2d-a}{2}=\frac{a+b+c+d}{2}\)

Vậy BĐT đc chứng minh