Question

[tống thị quỳnh]

cho a;b;c dương và a+b+c=1

cmr \(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge33\)

Answer 1

[alibaba nguyễn]

Ta chứng minh bổ đề:

\(\left(x+\frac{1}{x}\right)^2\ge\frac{260}{9}-\frac{160x}{3}\)

\(\Leftrightarrow\frac{9x^4+480x^3-242x^2+9}{9x^2}\ge0\)

\(\Leftrightarrow\frac{\left(3x-1\right)^2\left(x^2+54x+9\right)}{9x^2}\ge0\)(đúng)

Áp dụng vào bài toán ta được.

\(\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\)

\(\ge\frac{260}{9}-\frac{160a}{3}+\frac{260}{9}-\frac{160b}{3}+\frac{260}{9}-\frac{160c}{3}\)

\(=\frac{260}{3}-\frac{160}{3}\left(a+b+c\right)=\frac{260}{3}-\frac{160}{3}=\frac{100}{3}\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)

Answer 2

[Nguyễn Thiều Công Thành]

áp dụng bunhia ta có:

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

\(\ge\left(1+\frac{9}{a+b+c}\right)^2=100\)

\(\Rightarrow3\left[\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\right]\ge100\)

\(\Rightarrow\left(a+\frac{1}{a}\right)^2+\left(b+\frac{1}{b}\right)^2+\left(c+\frac{1}{c}\right)^2\ge\frac{100}{3}\left(Q.E.D\right)\)