Bạn chưa đăng nhập. Vui lòng đăng nhập để hỏi bài
Ngọc Huyền

Cho \(a^3\)+\(b^3\)+\(c^3\)=1. CM : \(\dfrac{a^2+b^2}{ab\left(a+b\right)^3}\)+\(\dfrac{b^2+c^2}{bc\left(b+c\right)^3}\)+\(\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\) \(\ge\) \(\dfrac{9}{4}\)

Kuro Kazuya
26 tháng 3 2017 lúc 12:31

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a^2+b^2}{ab\left(a+b\right)^3}\ge\dfrac{2ab}{ab\left(a+b\right)^3}=\dfrac{2}{\left(a+b\right)^3}\\\dfrac{b^2+c^2}{bc\left(b+c\right)^3}\ge\dfrac{2bc}{bc\left(b+c\right)^3}=\dfrac{2}{\left(b+c\right)^3}\\\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{2ca}{ca\left(c+a\right)^3}=\dfrac{2}{\left(c+a\right)^3}\end{matrix}\right.\)

\(\Rightarrow VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)

Chứng minh rằng \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{9}{8}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left\{{}\begin{matrix}2ab\le a^2+b^2\\2bc\le b^2+c^2\\2ca\le c^2+a^2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ab\le a^2-ab+b^2\\bc\le b^2-bc+c^2\\ca\le c^2-ca+a^2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}ab\left(a+b\right)\le\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\\bc\left(b+c\right)\le\left(b+c\right)\left(b^2-bc+c^2\right)=b^3+c^3\\ca\left(c+a\right)\le\left(c+a\right)\left(c^2-ca+a^2\right)=c^3+a^3\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3ab\left(a+b\right)\le3\left(a^3+b^3\right)\\3bc\left(b+c\right)\le3\left(b^3+c^3\right)\\3ca\left(c+a\right)\le3\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^3+3ab\left(a+b\right)+b^3\le4\left(a^3+b^3\right)\\b^3+3bc\left(b+c\right)+c^3\le4\left(b^3+c^3\right)\\c^3+3ca\left(c+a\right)+a^3\le4\left(c^3+a^3\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\left(a+b\right)^3\le4\left(a^3+b^3\right)\\\left(b+c\right)^3\le4\left(b^3+c^3\right)\\\left(c+a\right)^3\le4\left(c^3+a^3\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\left(a+b\right)^3}\ge\dfrac{1}{4\left(a^3+b^3\right)}\\\dfrac{1}{\left(b+c\right)^3}\ge\dfrac{1}{4\left(b^3+c^3\right)}\\\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4\left(c^3+a^3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\ge\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\)

Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức

\(\Rightarrow\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\ge\dfrac{9}{2\left(a^3+b^3+c^3\right)}=\dfrac{9}{2}\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{b^3+c^3}+\dfrac{1}{c^3+a^3}\right)\ge\dfrac{9}{8}\) ( đpcm )

Vậy \(2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\ge\dfrac{9}{4}\)

\(VT\ge2\left[\dfrac{1}{\left(a+b\right)^3}+\dfrac{1}{\left(b+c\right)^3}+\dfrac{1}{\left(c+a\right)^3}\right]\)

\(\Rightarrow VT\ge\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{ab\left(a+b\right)^3}+\dfrac{b^2+c^2}{bc\left(b+c\right)^3}+\dfrac{c^2+a^2}{ca\left(c+a\right)^3}\ge\dfrac{9}{4}\) ( đpcm )

Lightning Farron
26 tháng 3 2017 lúc 12:13

đề thiếu số dương à ? hay đủ


Các câu hỏi tương tự
lê thị tiều thư
Xem chi tiết
Ngọc Hiền
Xem chi tiết
lê thị tiều thư
Xem chi tiết
michelle holder
Xem chi tiết
michelle holder
Xem chi tiết
Neet
Xem chi tiết
Hàn Thiên Tử
Xem chi tiết
Quốc Bảo
Xem chi tiết
Nhâm Gia Nghĩa
Xem chi tiết