Question

Cho a,b,c dương thỏa: ab+bc+ca

Tìm min \(\frac{a^2}{\sqrt{a^3+8}}+\frac{b^2}{\sqrt{b^3+8}}+\frac{c^2}{\sqrt{c^3+8}}\\ \)

Answer 1

Another way: \(a+b+c\ge\sqrt{3\left(ab+bc+ac\right)}=3\)

Ta có BĐT phụ \(\frac{a^2}{\sqrt{a^3+8}}\ge\frac{11a}{18}-\frac{5}{18}\)

\(\Leftrightarrow\frac{\frac{\left(a-1\right)^2\left(121a^3-192a^2-480a+200\right)}{-324a^3-2592}}{\frac{a^2}{\sqrt{a^3+8}}+\frac{11a}{18}-\frac{5}{18}}\ge0\forall0< a\le1\)

TƯơng tự cho 2 BĐT còn lại ta cũng có:

\(\frac{b^2}{\sqrt{b^3+8}}\ge\frac{11b}{18}-\frac{5}{18};\frac{c^2}{\sqrt{c^3+8}}\ge\frac{11c}{18}-\frac{5}{18}\)

Cộng theo vế 3 BĐT trên ta có:

\(VT\ge\frac{11\left(a+b+c\right)}{18}-\frac{5}{18}\cdot3\ge1\)

"=" khi \(a=b=c=1\)

Answer 2

ab+bc+ac=3

Answer 3

\(a+b+c\ge\sqrt{3\left(ab+bc+ac\right)}=3\)

\(f\left(x\right)=\frac{x^2}{\sqrt{x^3+8}}\) là hàm lồi vì \(x>0\)

By Jensen'ineq: \(f\left(a\right)+f\left(b\right)+f\left(c\right)\ge3f\left(\frac{a+b+c}{3}\right)\ge3f\left(1\right)=1\)