Bài 1: Cho a+b=1. Tìm GTNN của biểu thức:
A = a(a2 + 2b) + b(b2 - a)
Bài 2: Cho tam giác có nửa chu vi p = \(\frac{a+b+c}{2}\)với a,b,c là độ dài ba cạnh.
CMR: \(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(1)\)
\(A=a\left(a^2+2b\right)+b\left(b^2-a\right)=a^3+2ab+b^3-ab=a^3+b^3+ab\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+ab=a^2+b^2\ge\frac{\left(a+b\right)^2}{1+1}=\frac{1}{2}\) ( Cauchy-Schwarz dạng Engel )
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=\frac{1}{2}\)
\(2)\)
\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}=\frac{1}{\frac{a+b+c}{2}-a}+\frac{1}{\frac{a+b+c}{2}-b}+\frac{1}{\frac{a+b+c}{2}-c}\)
\(=2\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)\)
Có : \(\hept{\begin{cases}b-a< c\\c-b< a\\a-c< b\end{cases}}\)
\(2\left(\frac{1}{-a+b+c}+\frac{1}{a-b+c}+\frac{1}{a+b-c}\right)>2\left(\frac{1}{2c}+\frac{1}{2a}+\frac{1}{2b}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) ???
1. A = a(a2 + 2b) + b(b2 - a)
A = a3 + 2ab + b3 - ab
A = a3 + ab + b3
A = ( a + b ) ( a2 - ab + b2 ) + ab
A = a2 + b2
Mà ( a - b )2 \(\ge\)0 với mọi a,b
\(\Rightarrow\)a2 + b2 \(\ge\)2ab \(\Rightarrow\)2 . ( a2 + b2 ) \(\ge\)( a + b )2 = 1 \(\Rightarrow\)( a2 + b2 ) \(\ge\)\(\frac{1}{2}\)
\(\Rightarrow\)A \(\ge\)\(\frac{1}{2}\) . Dấu " = " xảy ra \(\Leftrightarrow\)a = b \(\frac{1}{2}\)