Áp dụng BĐT Cauchy , có :
\(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge\frac{2\sqrt{a^3.b^3}}{ab}+\frac{2\sqrt{b^3.c^3}}{bc}+\frac{2\sqrt{c^3.a^3}}{ca}\)
\(\Leftrightarrow...........\ge2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\)
Lại có :
\(2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ca}\ge2a+2b+2c\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(b-2\sqrt{bc}+c\right)+\left(c-2\sqrt{ca}+c\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{b}-\sqrt{c}\right)^2+\left(\sqrt{c}-\sqrt{a}\right)^2\ge0\) (đúng)
Vậy \(\frac{a^3+b^3}{ab}+\frac{b^3+c^3}{bc}+\frac{c^3+a^3}{ca}\ge2a+2b+2c=2\left(a+b+c\right)\)
Ta có BĐT \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow\frac{a^3+b^3}{ab}\ge\frac{ab\left(a+b\right)}{ab}=a+b\)
Tương tự cũng có 2 BĐT:
\(\frac{b^3+c^3}{bc}\ge b+c;\frac{c^3+a^3}{ca}\ge c+a\)
Cộng theo vế được ĐPCM
Khi a=b=c
Ta có BĐT \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(\ge\left(a+b\right)\left(2ab-ab\right)=ab\left(a+b\right)\)
\(\Rightarrow\frac{a^3+b^3}{ab}\ge\frac{ab\left(a+b\right)}{ab}=a+b\)
Tương tự cũng có 2 BĐT:
\(\frac{b^3+c^3}{bc}\ge b+c;\frac{c^3+a^3}{ca}\ge c+a\)
Cộng theo vế được ĐPCM
Khi a=b=c
