Đặt M = (a^2+b^2-c^2)/2ab  + (b^2+c^2-a^2)/2bc + c^2+a^2-b^2/2ca

Ta có M-1=\(\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}-1.\)

=>M-1=\(\frac{c\left(a^2+b^2-c^2\right)+a\left(b^2+c^2-a^2\right)+b\left(c^2+a^2-b^2\right)-2abc}{2abc}\)

Vì a,b,c là độ dài 3 cạnh tam giác => a²+b²\(\ge\)c²,b²+c^2\(\ge\)a²,c²+a²\(\ge\)b²

Vậy M-1\(\ge\)0=> M\(\ge\)1(đpcm)

\(a^2+b^2+c^2=\left(a+b+c\right)^2\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)

\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)

\(\Leftrightarrow ab+ac+bc=0\)

\(\Leftrightarrow\hept{\begin{cases}ab=-ac-bc\\ac=-ab-bc\\bc=-ab-ac\end{cases}}\)

Ta có : \(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)

CMTT ta có : \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)

Thay vào A ta được :

\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)

\(A=\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{-a+c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=\frac{0}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(A=0\)

thiếu đề

bt \(M=\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2+b^2}{2ca}\)

Áp dụng BĐT Cauchy-SChwarz ta có:

\(VT=\frac{a^4}{a^2+2a^2bc}+\frac{b^4}{b^2+2ab^2c}+\frac{c^4}{c^2+2abc^2}\)

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

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2\cdot\frac{\left(ab+bc+ca\right)^2}{3}}\)

\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^2+b^2+c^2+2\cdot\frac{\left(a^2+b^2+c^2\right)^2}{3}}\)

\(\ge\frac{1^2}{1+2\cdot\frac{1^2}{3}}=\frac{3}{5}=VP\)

Dấu "=" bạn tự nghiên cứu nhé :D

DẤU BẰNG XẢY RA\(\Leftrightarrow a=b=c=\frac{1}{\sqrt{3}}\) CÁI NÀY LÀ ĐIỂM RƠI NHÉ.

tham khảo:        Câu hỏi của Nguyễn Thùy Trang     

https://olm.vn/hoi-dap/detail/240354680477.html

Cách : AM - GM :

\(VT=3-\left(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}\right)\left(1\right)\)

Áp dụng BĐT AM - GM :

\(\frac{2ab^2}{2ab^2+1}+\frac{2bc^2}{2bc^2+1}+\frac{2ca^2}{2ca^2+1}=\frac{2ab^2}{ab^2+ab^2+1}+\frac{2bc^2}{bc^2+bc^2+1}+\frac{2ca^2}{ca^2+ca^2+1}\)

\(\le\frac{2ab^2}{3\sqrt[3]{a^2b^4}}+\frac{2bc^2}{3\sqrt[3]{b^2c^4}}+\frac{2ca^2}{3\sqrt[3]{c^aa^4}}=\frac{2}{3}\left(\sqrt[3]{ab^2}+\sqrt[3]{bc^2}+\sqrt[3]{ca^2}\right)\)

\(\le\frac{2}{3}\left(\frac{a+b+b}{3}+\frac{b+c+c}{3}+\frac{c+a+a}{3}\right)=\frac{2}{3}\left(a+b+c\right)=2\left(2\right)\)

Từ (1) và (2) \(\Rightarrow VT\ge3-2=1\left(đpcm\right)\)