Xét TS

Có a^3 + b^3 + c^3 - 3abc = a^3 + 3a^2b + 3ab^2 + b^2 + c^3 - 3abc - 3a^2b - 3ab^2 = (a + b)^3 + c^3 - 3ab(a + b + c) = (a + b + c)(  (a+b)^2 + (a + b)c + c^2 - 3abc) = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ac) 

Rút gọn TS/MS được kết quả = a + b + c = 2009 => điều phải chứng minh

Ta có a³ + b³ + c³ - 3abc

=[ (a+ b)³ + c³ ] - [3ab(a+b) + 3abc]  = (a + b+ c)³ - 3(a + b).c(a + b + c) - 3ab.(a + b + c)

= (a + b+ c). [(a + b + c)² - 3c(a + b) - 3ab]

= (a + b+ c).(a² + b² + c² + 2ab + 2bc + 2ca - 3ac - 3bc - 3ab)

= (a + b + c)(a² + b² + c² - ab - bc - ca)

=> \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}=a+b+c=2009\)

Vậy.......

Ta có :

 \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\frac{\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\frac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\frac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\frac{\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)}{a^2+b^2+c^2-ab-ac-bc}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)}{a^2+b^2+c^2-ab-ac-bc}\)

\(=a+b+c=2009\)(đpcm)

\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-ba^3+ca^3-cb^3=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b+ab^2-c^3+a^2c+abc+b^2c\right)\)

\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(a-c\right)\left(b+c\right)\)

\(D=ab\left(a+b\right)+bc\left(b+c\right)+ac\left(c+a\right)+3abc=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ac\left(c+a\right)+abc=ab\left(a+b+c\right)+bc\left(a+b+c\right)++++ac\left(a+b+c\right)=\left(a+b+c\right)\left(ab+bc+ca\right)\)

D=ab(a+b)+bc(b+c)+ac(c+a)+3abc
= ab(a+b)+abc+bc(b+c)+abc+ac(c+a)+abc
= ab(a+b+c)+bc(b+c+a)+ac(c+a+b)
=( ab+bc+ac)(a+b+c)

Từ điều kiện ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3.\)

\(P=\frac{\frac{1}{a}}{1+\frac{b}{a}.\frac{c}{a}}+\frac{\frac{1}{b}}{1+\frac{c}{b}.\frac{a}{b}}+\frac{\frac{1}{c}}{1+\frac{a}{c}.\frac{b}{c}}\)

Đặt \(\left(\frac{1}{a};\text{ }\frac{1}{b};\text{ }\frac{1}{c}\right)=\left(x;y;z\right)\)

Thì \(x+y+z=3\)

\(P=\frac{x}{1+\frac{z}{x}}+\frac{y}{1+\frac{x}{y}}+\frac{z}{1+\frac{y}{z}}=\frac{x^2}{x+z}+\frac{y^2}{y+x}+\frac{z^2}{z+y}\ge\frac{\left(x+y+z\right)^2}{x+z+y+x+z+y}=\frac{\left(x+y+z\right)^2}{2}=\frac{9}{2}.\)

bạn còn cách nào khác như biến đổi thẳng luôn trong vế trái thay vì đặt x,y ,z được không ? Cảm ơn nhiều !

ban co si duoi mau . sau do ap dung co si nguoc cho ca phan so la ra

Câu hỏi của TRẦN HỮU ĐẠT - Toán lớp 9 - Học toán với OnlineMath

1) \(\Sigma\frac{a}{b^3+ab}=\Sigma\left(\frac{1}{b}-\frac{b}{a+b^2}\right)\ge\Sigma\frac{1}{a}-\Sigma\frac{1}{2\sqrt{a}}=\Sigma\left(\frac{1}{a}-\frac{2}{\sqrt{a}}+1\right)+\Sigma\frac{3}{2\sqrt{a}}-3\)

\(\ge\Sigma\left(\frac{1}{\sqrt{a}}-1\right)^2+\frac{27}{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}-3\ge\frac{27}{2\sqrt{3\left(a+b+c\right)}}-3=\frac{3}{2}\)

2.

Vỉ \(ab+bc+ca+abc=4\)thi luon ton tai \(a=\frac{2x}{y+z};b=\frac{2y}{z+x};c=\frac{2z}{x+y}\)

\(\Rightarrow VT=2\Sigma_{cyc}\sqrt{\frac{ab}{\left(b+c\right)\left(c+a\right)}}\le2\Sigma_{cyc}\frac{\frac{b}{b+c}+\frac{a}{c+a}}{2}=3\)

Cho o dong 2 la x,y,z nhe,ghi nham