Question

[Pham Van Hung]

Cho a;b;c > 0 thỏa mãn \(a+b+c\le\frac{3}{2}\)

Tìm GTNN của \(P=\left(3+\frac{1}{a}+\frac{1}{b}\right)\left(3+\frac{1}{b}+\frac{1}{c}\right)\left(3+\frac{1}{c}+\frac{1}{a}\right)\)

Answer 1

[tth_new]

\(P=\left[\left(2+\frac{1}{a}+\frac{1}{b}\right)+1\right]\left[\left(2+\frac{1}{b}+\frac{1}{c}\right)+1\right]\left[\left(2+\frac{1}{c}+\frac{1}{a}\right)+1\right]\)

\(\ge\left(6\sqrt[3]{\frac{1}{4ab}}+1\right)\left(6\sqrt[3]{\frac{1}{4bc}}+1\right)\left(6\sqrt[3]{\frac{1}{4ca}}+1\right)\)

\(\ge\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ab}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4bc}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ca}}\right)^6}\right]\)

\(=\left[7\sqrt[7]{\left(\frac{1}{4ab}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4bc}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4ca}\right)^2}\right]\)

\(=343\sqrt[7]{\left(\frac{1}{64\left(abc\right)^2}\right)^2}\ge343\sqrt[7]{\left(\frac{1}{64\left[\frac{\left(a+b+c\right)^3}{27}\right]^2}\right)^2}=343\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

P/s: Em chưa check lại đâu nha::D

Answer 2

[tth_new]

Khúc cuối bài ban nãy là \(\ge343\) nha! Em đánh nhầm

Cách khác (em thử dùng Holder, mới học nên em không chắc lắm):

\(P\ge\left(3+\sqrt[3]{\frac{1}{abc}}+\sqrt[3]{\frac{1}{abc}}\right)^3=\left(3+2\sqrt[3]{\frac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\frac{1}{\left[\frac{\left(a+b+c\right)^3}{27}\right]}}\right)^3\ge343\)