1/ Ta có \(a^3+b^3\ge ab\left(a+b\right)\)
Thật vậy, BĐT tương đương:
\(a^3-a^2b+b^3-ab^2\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\) (luôn đúng)
\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b\)
2/ \(P=\frac{a^2}{ab+ac}+\frac{b^2}{bc+bd}+\frac{c^2}{cd+ca}+\frac{d^2}{ad+bd}\ge\frac{\left(a+b+c+d\right)^2}{2ac+2bd+ab+bc+cd+ad}\)
\(P\ge\frac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+ab+bc+cd+ad}\)
\(P\ge\frac{4ac+4bd+2ab+2bc+2cd+2ad}{2ac+2bd+ab+bc+cd+ad}=2\)
Dấu "=" xảy ra khi \(a=b=c=d\)
