Question

[Kim Yuri]

Cho a, b \(\ge\)0. CMR: a³+b³\(\ge\)ab(a + b)

Áp dụng CM: \(\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{a^3+c^3+1}\le1\) với a, b, c > 0 thỏa mãn abc=1

Answer 1

[Võ Hồng Phúc]

\(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0,\forall a,b\ge0\)

Áp dụng:

\(\frac{1}{a^3+b^3+1}\le\frac{1}{ab\left(a+b\right)+1}=\frac{abc}{ab\left(a+b\right)+abc}=\frac{c}{a+b+c}\)

\(\frac{1}{b^3+c^3+1}\le\frac{1}{bc\left(b+c\right)+1}=\frac{abc}{bc\left(b+c\right)+abc}=\frac{a}{a+b+c}\)

\(\frac{1}{c^3+a^3+1}\le\frac{1}{ca\left(c+a\right)+1}=\frac{abc}{ca\left(c+a\right)+abc}=\frac{b}{a+b+c}\)

\(\Rightarrow VT\le\frac{c}{a+b+c}+\frac{a}{a+b+c}+\frac{b}{a+b+c}=\frac{a+b+c}{a+b+c}=1\left(đpcm\right)\)