cho a,b,c>0 CMR:
a.[tex]a^3b+b^3c+c^3a\geq a^2bc+b^2ac+c^2ab[/tex]
b.[tex]\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\geq \frac{a+b+c}{3}[/tex]
Câu b)
Ta có [tex]a-b=\frac{a^3-b^3}{a^2+ab+b^2} (1)[/tex]
[tex]b-c=\frac{b^3-c^3}{b^2+bc+c^2} (2)[/tex]
[tex]c-a=\frac{c^3-a^3}{c^2+ac+a^2} (3)[/tex]
Cộng theo từng vế [tex] (1), (2), (3)[/tex], ta có
[tex]\frac{a^3-b^3}{a^2+ab+b^2}+\frac{b^3-c^3}{a^2+ab+b^2}+\frac{c^3-a^3}{a^2+ac+c^2}=0[/tex]
[tex]=> \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{a^2+ac+c^2}[/tex]=[tex]\frac{1}{2}(\frac{a^3+b^3}{a^2+ab+b^2}+\frac{b^3+c^3}{a^2+ab+b^2}+\frac{c^3+a^3}{a^2+ac+c^2})[/tex]
Đặt S= [tex]=> \frac{a^3}{a^2+ab+b^2}+\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{a^2+ac+c^2}[/tex]
Ta có [tex]2(x-y)^2\geq 0 => 2(x^2-2xy+y^2)\geq 0 => 3(x^2-xy+y^2)\geq x^2+xy+y^2[/tex]
=> [tex]3(x^2-xy+y^2)(x+y)\geq (x^2+xy+y^2)(x+y)[/tex]
=> [tex]\frac{x^3+y^3}{x^2+xy+y^2}\geq \frac{1}{3}(x+y)[/tex]
Vậy ta có [tex]\frac{a^3+b^3}{a^2+ab+b^2}\geq \frac{1}{3}(a+b)[/tex]
Tương tự thế => [tex]2S\geq \frac{2}{3}(a+b+c) => S\geq \frac{a+b+c}{3}[/tex] (đpcm)
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
