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\(\frac{1}{a^2}=\frac{1}{\left(bc\right)^2}\)
\(\Rightarrow\frac{1}{a^2}+1=\frac{1}{\left(bc\right)^2}+1\ge2\frac{1}{bc}=2a\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow a+b+c=0\)
Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Đặt A =\(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\)
Vì a + b \(\ne\)0 nên A luôn được xác định.
Giả sử \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\)
\(\Leftrightarrow\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)^2}+\frac{\left(ab+1\right)^2}{\left(a+b\right)^2}-\frac{2\left(a+b\right)^2}{\left(a+b\right)^2}\ge0\)
\(\Leftrightarrow\left(a^2+b^2\right)\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)(vì a + b \(\ne\)0)
\(\Leftrightarrow[\left(a^2+2ab+b^2\right)-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow[\left(a+b\right)^2-2ab]\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^4-2ab\left(a+b\right)^2+\left(ab+1\right)^2-2\left(a+b\right)^2\ge0\)
\(\Leftrightarrow\left(a+b\right)^4-\left[2ab\left(a+b\right)^2+2\left(a+b\right)^2\right]+\left(ab+1\right)^2\ge0\)
\(\Leftrightarrow\left[\left(a+b\right)^2\right]^2-2\left(a+b\right)^2\left(ab+1\right)+\left(ab+1\right)^2\ge0\)
\(\left[\left(a+b\right)^2-\left(ab+1\right)^2\right]^2\ge0\)(luôn đúng)
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a+b\ne0\\\Leftrightarrow a=b\end{cases}}\Leftrightarrow a=b\left(a,b\ne0\right)\)
Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge\)2 với a, b là các số thỏa mãn a+b \(\ne\)0
Dấu bằng xảy ra
\(\Leftrightarrow\hept{\begin{cases}a=b\\a+b\ne0\end{cases}\Leftrightarrow a=b}\)(a,b \(\ne\)0)
Vậy \(a^2+b^2+\left(\frac{ab+1}{a+b}\right)^2\ge2\) với a, b là các số thỏa mãn \(a+b\ne0\)
\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
...
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c+b+a}{abc}\right)\)
Mà a+b+c = 0 nên suy ra:
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{0}{abc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
Ta có: (\(\frac{1}{a}\)+\(\frac{1}{b}\)+\(\frac{1}{c}\))\(^2\)= \(\frac{1}{a^2}\)+\(\frac{1}{b^2}\)+\(\frac{1}{c^2}\)+\(\frac{2}{abc}\)(\(\frac{a+b+c}{abc}\))
Mà
A+B+C= 0
nên: VT = VP (đpcm)
