Có: \(a^2=b^2+c^2\)
\(\Leftrightarrow a^2=\left(b+c\right)^2-2bc\)
\(\Leftrightarrow\sqrt{a^2}=\sqrt{\left(b+c\right)^2-2bc}\)
\(\Leftrightarrow a=\sqrt{\left(b+c\right)^2-2bc}< \sqrt{\left(b+c\right)^2}=b+c\)
Vậy a<b+c
Câu kia pt hdt thức
Có: \(b^3+c^3=\left(b+c\right)\left(b^2+c^2-bc\right)=\left(b+c\right)\left(a^2-bc\right)\)
Theo câu a) a<b+c
Nên: \(b^3+c^3< a\left(a^2-bc\right)=a^3-abc< a^3\)
Vậy a^3>b^3+c^3