\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{ab}{2b}\right)\)

\(=\dfrac{1}{9}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Tương tự:

\(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}+\dfrac{b}{2}\right)\)

\(\dfrac{ac}{c+3a+2b}\le\dfrac{1}{9}\left(\dfrac{ac}{b+c}+\dfrac{ac}{a+b}+\dfrac{c}{2}\right)\)

Cộng vế:

\(P\le\dfrac{1}{9}\left(\dfrac{bc+ac}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ab+ac}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(P\le\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

tham hảo nha:

https://hoidap247.com/cau-hoi/24991

# mui #

a) \(\left(x+2y\right)^2=x^2+2.x.2y+\left(2y\right)^2=x^2+4xy+4y^2\)

b) \(\left(3-x\right).\left(3+x\right)=9+3x-3x-x^2=9-x^2=3^2-x^2\)

c) \(\left(5-x\right)^2=5^2-2.5.x+x^2=25-10x+x^2\)

d) \(\left(3+y\right)^2=3^2+2.3.y+y^2=9+6y+y^2\)

a) x²+4xy+4y² b)x(9-x²) = 9x-x³ c)25-10x+x² d)9+6y+x²

​

\(ab+bc+ca=abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

\(\dfrac{1}{a+2b+3c}=\dfrac{1}{a+b+b+c+c+c}\le\dfrac{1}{6^2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}\right)\)

\(\Rightarrow\dfrac{1}{a+2b+3c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{3}{c}\right)\)

Tương tự:

\(\dfrac{1}{b+2c+3a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{2}{c}+\dfrac{3}{a}\right)\) ; \(\dfrac{1}{c+2a+3b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{2}{a}+\dfrac{3}{b}\right)\)

Cộng vế:

\(P\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{2bc}{2bc+a^2}+\dfrac{2ac}{2ac+b^2}+\dfrac{2ab}{2ab+c^2}\le2\)

\(\Leftrightarrow\dfrac{2bc}{2bc+a^2}-1+\dfrac{2ac}{2ac+b^2}-1+\dfrac{2ab}{2ab+c^2}-1\le2-3\)

\(\Leftrightarrow\dfrac{a^2}{2bc+a^2}+\dfrac{b^2}{2ac+b^2}+\dfrac{c^2}{2ab+c^2}\ge1\)

BĐT trên đúng theo C-S:

\(\dfrac{a^2}{2bc+a^2}+\dfrac{b^2}{2ac+b^2}+\dfrac{c^2}{2ab+c^2}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=1\)

Dấu "=" xảy ra khi \(a=b=c\)

cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)

\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)

\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)

\(=x-1=2013-1=2012\)

nhiều quá

lớp 7 hả

Bài toán cơ bản:

\(abc=1\Rightarrow\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}=1\) 

Bunhiacopxki:

\(\left(a+b+c\right)\left(\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\right)\ge\left(\dfrac{a}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{c}{ac+c+1}\right)^2=1\)

\(\Rightarrow\dfrac{a}{\left(ab+a+1\right)^2}+\dfrac{b}{\left(bc+b+1\right)^2}+\dfrac{c}{\left(ac+c+1\right)^2}\ge\dfrac{1}{a+b+c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

Cách 2:

Do \(abc=1\), đặt \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

Ta có \(\dfrac{a}{\left(ab+a+1\right)^2}=\dfrac{\dfrac{x}{y}}{\left(\dfrac{x}{z}+\dfrac{x}{y}+1\right)^2}=\dfrac{\dfrac{x}{y}.y^2z^2}{\left(xy+yz+zx\right)^2}=\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}\)...

Từ đó, BĐT cần chứng minh trở thành:

\(\dfrac{xyz^2}{\left(xy+yz+zx\right)^2}+\dfrac{x^2yz}{\left(xy+yz+zx\right)^2}+\dfrac{xy^2z}{\left(xy+yz+zx\right)^2}\ge\dfrac{1}{\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}}\)

\(\Leftrightarrow xyz\left(x+y+z\right)\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)\ge\left(xy+yz+zx\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2z+y^2x+z^2y\right)\ge\left(xy+yz+zx\right)^2\)

Thật vậy, áp dụng BĐT Bunhiacopxki:

\(\left(z+x+y\right)\left(x^2z+y^2x+z^2y\right)\ge\left(\sqrt{zx^2z}+\sqrt{xy^2x}+\sqrt{yz^2y}\right)^2=\left(xy+yz+zx\right)^2\) (đpcm)