View attachment 125294
Mn giúp e với ạ
E đg cần gấp
Mai e phải nộp bài cho cô rồi ạ
((((
bài 1:
[tex]\frac{a^5}{a^2+ab+b^2}+\frac{b^5}{b^2+bc+c^2}+\frac{c^5}{c^2+ca+a^2}\\\\ =\frac{a^6}{a^3+a^2b+ab^2}+\frac{b^6}{b^3+b^2c+bc^2}+\frac{c^5}{c^3+a^2c+ac^2}\\\\ \geq \frac{(a^3+b^3+c^3)^2}{a^3+b^3+c^3+a^2b+ab^2+b^2c+bc^2+a^2c+ac^2}\\\\ +, a^3+b^3=(a+b).(a^2+b^2-ab)\\\\ \geq (a+b).(2ab-ab)=(a+b).ab=a^2b+ab^2\\\\ +,b^3+c^3\geq b^2c+bc^2\\\\ +,c^3+a^3\geq a^2c+ac^2\\\\ => a^2b+ab^2+b^2c+bc^2+a^2c+ac^2\leq 2.(a^3+b^3+c^3)\\\\ => đpcm[/tex]
bài 2:
[tex]\frac{x}{x^2-yz+2016}+\frac{y}{y^2-xz+2016}+\frac{z}{z^2-xy+2016}\\\\ =\frac{x^2}{x^3-xyz+2016x}+\frac{y^2}{y^3-xyz+2016y}+\frac{z^2}{z^3-xyz+2016z}\\\\ \geq \frac{(x+y+z)^2}{(x+y+z).(x^2+y^2+z^2-xy-yz-xz+2016)}\\\\ =\frac{(x+y+z)^2}{(x+y+z).(x^2+y^2+z^2+2xy+2yz+2xz)}\\\\ =\frac{1}{x+y+z}[/tex]
bài 3:
[tex]\frac{1+3a}{b^2+1}+\frac{1+3b}{c^2+1}+\frac{1+3c}{a^2+1}\\\\ =\frac{1+a}{b^2+1}+\frac{1+b}{c^2+1}+\frac{1+c}{a^2+1}+2.(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1})\\\\ +, \frac{a}{b^2+1}=a-\frac{ab^2}{b^2+1}\geq a-\frac{ab^2}{2b}=a-\frac{ab}{2}\\\\ => \frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\geq a+b+c-\frac{3}{2}\\\\ +,\frac{a+1}{b^2+1}=a+1-\frac{(a+1).b^2}{b^2+1}\geq a+1-\frac{(a+1).b^2}{2b}\\\\ =a+1-\frac{ab+b}{2}\\\\ =>\frac{1+a}{b^2+1}+\frac{1+b}{c^2+1}+\frac{1+c}{a^2+1}\geq 3+\frac{a+b+c}{2}-\frac{3}{2}\\\\ +, (a+b+c)^2\geq 3.(ab+bc+ca)=9\\\\ => a+b+c\geq 3\\\\ =>....[/tex]
bài 4:
[tex]\frac{x^3}{x^3+3yzt}+\frac{y^3}{y^3+3xzt}+\frac{z^3}{z^3+3xyt}+\frac{t^3}{t^3+3xyz}\\\\ \geq \frac{x^3}{x^3+y^3+z^3+t^3}+\frac{y^3}{x^3+y^3+z^3+t^3}+\frac{z^3}{x^3+y^3+z^3+t^3}+\frac{t^3}{x^3+y^3+z^3+t^3}\\\\ =1\\\\ (y^3+z^3+t^3\geq 3yzt)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
