a³ + b³=(a+b)(a²-ab+b²)

(a + b)³ =a³+b³+3ab(a+b)

a² + b²=a²+2ab+b²

SGK TOÁN 8 TẬP 1

\(\frac{1}{\left(b-c\right)\left(a^2+ac-b^2-bc\right)}+\frac{1}{\left(c-a\right)\left(b^2+ab-c^2-ac\right)}-\frac{1}{\left(a-b\right)\left(a^2+ab-c^2-ac\right)}\)

\(=\frac{1}{\left(b-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]}+\frac{1}{\left(c-a\right)\left[\left(b-c\right)\left(b+c\right)+a\left(b-c\right)\right]}-\frac{1}{\left(a-b\right)\left[\left(a-c\right)\left(a+c\right)-b\left(a-c\right)\right]}\)

\(=\frac{c-a}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{a-b}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}-\frac{b-c}{\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a+b+c\right)}\)

\(=\frac{c-a}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{a-b}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{b-c}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)

\(=\frac{0}{\left(a-b\right)\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}\)

\(=0\)

\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)

\(=\left[\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-4y^2\)

\(=\left(x+y-x+y\right)^2-\left(2y\right)^2\)

\(=\left(2y\right)^2-\left(2y\right)^2=0\)

Sửa:

\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)

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

\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)

\(=2x^2-6y^2+4xy\)

\(=2\left(x^2-3y^2+2xy\right)\)

\(=2\left(x^2-3y^2+3xy-xy\right)\)

\(=2\left[x\left(x-y\right)+3y\left(x-y\right)\right]\)

\(=2\left(x-y\right)\left(3y+x\right)\)

x ở đâu ra vại @

BĐT \(\Leftrightarrow2\left(a^2+b^2+c^2\right)\ge a+b+c+ab+bc+ca\)

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{1}{4}\left(y+z-x\right)^2+a^2+b^2+c^2-\left(a+b+c\right)\ge0\)

Có: \(VT\ge\frac{3}{4}\left(y-z\right)^2+\frac{1}{4}\left(y+z-x\right)^2+\left[\frac{\left(a+b+c\right)^2}{3}-\left(a+b+c\right)\right]\ge0\)(chú ý: \(\left(a+b+c\right)^2=\left(a+b+c\right)\left(a+b+c\right)\ge3\sqrt[3]{abc}\left(a+b+c\right)=3\left(a+b+c\right)\))

Ta có đpcm.

Có cách khác ^_^ mới nghĩ ra

BĐt <=> \(P\left(a,b,c\right)=a^2+b^2+c^2-\frac{1}{2}\left(a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge0\)

Không mất tính tổng quát , giả sử : \(a=min\left\{a,b,c\right\}\Rightarrow t=\sqrt{bc}\ge1\)

=> Chứng minh: \(P\left(a,b,c\right)\ge P\left(a,t,t\right)\)

Thật vậy , \(P\left(a,b,c\right)-P\left(a,t,t\right)=\left(\sqrt{b}-\sqrt{c}\right)^2\left[\left(\sqrt{b}+\sqrt{c}\right)^2-\frac{1}{2}\left(1+\frac{1}{bc}\right)\right]\)

                                                                  \(\ge\left(\sqrt{b}-\sqrt{c}\right)^2\left[4-\frac{1}{2}\left(1+1\right)\right]\ge0\)

mặt khác: \(P\left(a,t,t\right)=P\left(\frac{t}{t^2},t,t\right)=\frac{\left(t-1\right)^2\left(3t^4+4t^3+5t^2+4t+2\right)}{2t^4}\ge0\)

=> BĐT được chứng minh . Đt xảy ra<=> a=b=c=1