Question

Cho a, b, c là 3 số dương thỏa mãn abc=1 . Chứng minh rằng :

1/a³(b+c) + 1/b³(c+a) +1/c³(a+b) \(\ge\) 3/2.

Các bn giúp mk vs. Mk cần gấp huhu.

Answer 1

Đặt \(\left(a;b;c\right)=\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\) \(\Rightarrow xyz=1\)

\(P=\frac{1}{\frac{1}{x^3}\left(\frac{1}{y}+\frac{1}{z}\right)}+\frac{1}{\frac{1}{y^3}\left(\frac{1}{z}+\frac{1}{x}\right)}+\frac{1}{\frac{1}{z^3}\left(\frac{1}{x}+\frac{1}{y}\right)}\)

\(P=\frac{x^3yz}{y+z}+\frac{y^3xz}{x+z}+\frac{z^3xy}{x+y}=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)

\(\Rightarrow P\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{1}{2}\left(x+y+z\right)\ge\frac{1}{2}.3\sqrt[3]{xyz}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(x;y;z\right)=\left(1;1;1\right)\)