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Lời giải:
Vì \(ax^2+bx+c=0\) vô nghiệm nên \(\Delta=b^2-4ac< 0\)
\(\Rightarrow b^2< 4ac\)
Kết hợp với \(a,b>0\Rightarrow c>0\)
Theo BĐT Cô-si: \(4\sqrt{ac}\leq 4a+c\Rightarrow 4ac\leq \frac{(4a+c)^2}{4}\)
Do đó: \(b^2< \frac{(4a+c)^2}{4}\Rightarrow (2b)^2< (4a+c)^2\). Với \(a,b,c>0\)
\(\Rightarrow 2b< 4a+c\)
\(\Rightarrow a+b+c> 3(b-a)\)
Mà: \(b-a>0\Rightarrow \frac{a+b+c}{b-a}> \frac{3(b-a)}{b-a}=3\) (đpcm)
1) \(a+b+c=0\Rightarrow2\left(a+b+c\right)=0\Rightarrow\frac{2\left(a+b+c\right)}{abc}=0\)
\(\Rightarrow M=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(\Rightarrow M=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{yz}+\frac{2}{zx}+\frac{2}{xy}\)
\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}\)
\(=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c+b+a}{abc}\right)\)
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)
Đặt \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\) ; \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
Ta có : \(P=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)}{abc}\)
Xét tử số của P : \(ab\left(a-b\right)+bc\left(b-c\right)+ac\left(c-a\right)=ab\left[-\left(b-c\right)-\left(c-a\right)\right]+bc\left(b-c\right)+ac\left(c-a\right)\)
\(=-ab\left(b-c\right)-ab\left(c-a\right)+bc\left(b-c\right)+ac\left(c-a\right)\)
\(=b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(c-b\right)=\left(b-c\right)\left(c-a\right)\left(b-a\right)\)
\(\Rightarrow P=\frac{\left(b-c\right)\left(c-a\right)\left(b-a\right)}{abc}\)
Lại có : \(Q=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\). Đặt \(a-b=x\); \(b-c=y\); \(c-a=z\)
Suy ra được : \(\hept{\begin{cases}x-y=a-b-b+c=a+c-2b=-3b\\y-z=b-c-c+a=a+b-2c=-3c\\z-x=c-a+b-a=b+c-2a=-3a\end{cases}\Rightarrow\hept{\begin{cases}b=-\frac{\left(x-y\right)}{3}\\c=-\frac{\left(y-z\right)}{3}\\a=-\frac{\left(z-x\right)}{3}\end{cases}}}\)
Ta có : \(Q=\frac{-\left(\frac{y-z}{3}\right)}{x}+\frac{-\left(\frac{z-x}{3}\right)}{y}+\frac{-\left(\frac{x-y}{3}\right)}{z}=-\frac{1}{3}.\left(\frac{y-z}{x}+\frac{z-x}{y}+\frac{x-y}{z}\right)\)
\(=-\frac{1}{3}\left(\frac{yz\left(y-z\right)+xz\left(z-x\right)+yx\left(x-y\right)}{xyz}\right)\)
Đến đây rút gọn tương tự với P được: \(Q=\frac{\left(x-z\right)\left(x-y\right)\left(z-y\right)}{3xyz}=\frac{\left(3a\right).\left(-3b\right).\left(3c\right)}{3\left(a-b\right)\left(b-c\right)\left(c-a\right)}\Rightarrow Q=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
Vậy : \(PQ=\frac{\left(b-c\right)\left(c-a\right)\left(b-a\right)}{abc}.\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=9\)
Vậy ta có điều phải chứng minh.
\(\)
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}\right)^2+\left(\frac{1}{b}\right)^2+\left(\frac{1}{c}\right)^2+2\frac{1}{ab}+2\frac{1}{bc}+2\frac{1}{ac}\)
\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}\)
\(\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=0\\ 2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=0\)
\(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=0\\ \frac{abc^2+a^2bc+ab^2c}{a^2b^2c^2}=0\)
\(abc^2+a^2bc+ab^2c=0\\ abc\left(c+a+b\right)=0\)
\(a+b+c=0\)(DPCM)
Do phương trình \(ax^2+bx+c\)vô nghiệm nên ta có:
\(b^2-4ac< 0\)
\(\Leftrightarrow4ac>b^2\)
Mà \(b>a>0\)
\(\Rightarrow c>0\)
Giả sử \(\frac{a+b+c}{b-a}>3\) \(\left(1\right)\)
\(\Leftrightarrow a+b+c>3b-3a\)
\(\Leftrightarrow4a+c>2b\)
Lại có: \(\left(4a+c\right)^2\ge16ac>4b^2\)
\(\Rightarrow4a+c>2b\)
Suy ra (1) đúng.
Vậy \(\frac{a+b+c}{b-a}>3\)