Question

1, cho biết: x=by+cz ; y =ax+cz ; z=ax+by. CMR \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2\)                                                                                          2, Cho a,b,c >0 và abc=1 Tìm GTLN của biểu thức P= \(\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ac}{c^5+a^5+ac}\)

Answer 1

B1:Cong 7heo ve cac gia 7hie7: \(x+y+z=2\left(ax+by+cz\right)\)

Ma` \(x=by+cz\Leftrightarrow x\left(a+1\right)=ax+by+cz=\frac{x+y+z}{2}\)

\(\Leftrightarrow\frac{1}{a+1}=\frac{2x}{x+y+z}\).7uong 7u cho 2 dang 7huc con lai roi cong 7heo ve:

\(V7=\frac{2\left(x+y+z\right)}{x+y+z}=2=VP\) (DPCM)

B2: chu y \(a^5+b^5=\left(a+b\right)\left(a^4-a^3b+a^2b^2-ab^3+b^4\right)\)

\(=\left(a+b\right)\left(a^3\left(a-b\right)+a^2b^2-b^3\left(a-b\right)\right)\)

\(=\left(a+b\right)\left(\left(a-b\right)\left(a^3-b^3\right)+a^2b^2\right)\)

\(=\left(a+b\right)\left(\left(a-b\right)^2\left(a^2+b^2-ab\right)+a^2b^2\right)\)

\(\ge ab\left(a+b\right)\left(a^2+b^2-ab\right)\)\(\ge a^2b^2\left(a+b\right)\)

\(\Leftrightarrow a^5+b^5+ab\ge ab\left(ab\left(a+b\right)+abc\right)=a^2b^2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{a^5+b^5+ab}\le\frac{abc}{ab\left(a+b+c\right)}=\frac{c}{a+b+c}\)

7uong 7u cho 2 BD7 con lai roi cong 7heo ve

\(V7\le\frac{a+b+c}{a+b+c}=1=VP\)

Dau "=" khi \(a=b=c=1\)