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Bài 1:
\(\frac{ab}{(a-c)(b-c)}+\frac{bc}{(b-a)(c-a)}+\frac{ca}{(c-b)(a-b)}=\frac{-ab}{(c-a)(b-c)}+\frac{-bc}{(a-b)(c-a)}+\frac{-ca}{(b-c)(a-b)}\)
\(=\frac{-ab(a-b)}{(a-b)(b-c)(c-a)}+\frac{-bc(b-c)}{(a-b)(b-c)(c-a)}+\frac{-ca(c-a)}{(a-b)(b-c)(c-a)}\)
\(=\frac{-ab(a-b)-bc(b-c)-ca(c-a)}{(a-b)(b-c)(c-a)}=\frac{-(a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)}{-(a^2b+b^2c+c^2a)+(ab^2+bc^2+ca^2)}=1\)
Bài 2:
Ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow \frac{a+b}{ab}+\frac{a+b}{c(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{c(a+b+c)+ab}{abc(a+b+c)}=0\)
\(\Leftrightarrow (a+b).\frac{(c+a)(c+b)}{abc(a+b+c)}=0\Rightarrow (a+b)(b+c)(c+a)=0\)
\(\Rightarrow \left[\begin{matrix} a+b=0\\ b+c=0\\ c+a=0\end{matrix}\right.\)
Không mất tổng quát giả sử $a+b=0$
Khi đó:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3}+\frac{1}{(-a)^3}+\frac{1}{c^3}=\frac{1}{c^3}(1)\)
\(\frac{1}{a^3+b^3+c^3}=\frac{1}{a^3+(-a)^3+c^3}=\frac{1}{c^3}(2)\)
Từ \((1);(2)\Rightarrow \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^3+b^3+c^3}\) (đpcm)
Ta có: abc = 1, thế vào ta được:
\(\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)
\(=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(c+a\right)}+\frac{ab}{c^2\left(a+b\right)}\)
\(=\frac{b^2c^2}{a^2bc\left(b+c\right)}+\frac{c^2a^2}{b^2ac\left(c+a\right)}+\frac{a^2b^2}{c^2ab\left(a+b\right)}\)
Áp dụng BĐT Cauchy - Schwarz dạng Engel, ta có:
\(VT\ge\frac{\left(bc+ca+ac\right)^2}{abc\left(2ab+2bc+2ca\right)}=\frac{\left(bc+ca+ac\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{\sqrt[3]{a^2b^2c^2}}{2}=\frac{3}{2}\)
\("="\Leftrightarrow a=b=c=1\)
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)
\(\Leftrightarrow\left(\frac{1}{a-b}+\frac{1}{c-a}+\frac{1}{b-c}\right).\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a}{\left(a-b\right)\left(b-c\right)}+\frac{a}{\left(c-a\right)\left(b-c\right)}+\frac{b}{\left(c-a\right)\left(a-b\right)}+\frac{b}{\left(c-a\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(b-c\right)}+\frac{c}{\left(a-b\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{a\left(c-a\right)+a.\left(a-b\right)+b.\left(a-b\right)+b.\left(b-c\right)+c.\left(b-c\right)+c.\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+\frac{ac-a^2+ab-ac+ba-b^2+b^2-bc+bc-c^2+c^2-ac}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}+0=0\)
\(\Leftrightarrow\frac{a}{\left(b-c\right)^2}+\frac{b}{\left(a-c\right)^2}+\frac{c}{\left(a-b\right)^2}=0\)
đpcm
Bài làm :
Ta có :
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ac=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Leftrightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\left(1\right)\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\left(2\right)\)
Thay (1) vào (2) ; ta được :
\(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
=> Điều phải chứng minh
Ta có \(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2ac+2bc=0\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
Ta lại có giả sử
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=3\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3.a^2b^2c^2\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3-3.a^2b^2c^2=0\)
\(\Leftrightarrow\left(ab+bc+ac\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ac\right)-3ab^3c\left(-ac\right)-3a^2b^2c^2=0\)
\(\Leftrightarrow0+3a^2b^2c^2-3a^2b^2c^2+0=0\)
\(\Leftrightarrow0=0\left(lđ\right)\)
Vậy bất đẳng thức được chứng minh
\(VT=\frac{c-b}{\left(a-b\right)\left(c-a\right)}+\frac{a-c}{\left(a-b\right)\left(b-c\right)}+\frac{b-a}{\left(b-c\right)\left(c-a\right)}\)
\(=\frac{-\left(b-c\right)^2-\left(c-a\right)^2-\left(a-b\right)^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{-2a^2-2b^2-2c^2+2ab+2ac+2bc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2ab-2ac+2bc-2b^2+2ab+2ac-2bc-2a^2-2ab+2ac+2bc-2c^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2\left(a-b\right)\left(b-c\right)+2\left(a-b\right)\left(c-a\right)+2\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{2}{c-a}+\frac{2}{b-c}+\frac{2}{a-b}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\frac{yz+xz+xy}{xyz}=0\)
yz + xz + xy = 0
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)
a) Từ giả thiết suy ra: xy + yz + zx = 0
Do đó:
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)
b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)
Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra điều phải chứng minh
Cách 1:
Sau khi phân tích hoàn toàn ra , ta cần CM
\(Σ\left(a^4b+a^4c+3a^3b^2+3a^3c^2-8a^2b^2c\right)\ge0\)
Đúng theo BĐT Muirhead
Cách 2: Giải theo SOS, ta cần CM
\(\frac{\left(a+b+c\right)^2}{6abc}\geΣ\frac{1}{a+b}\) tức là \(\frac{\left(a+b+c\right)^3}{3abc}\ge2Σ\frac{a+b+c}{a+b}\)
tức là \(\frac{\left(a+b+c\right)^3}{3abc}\ge6+Σ\frac{2c}{a+b}\) tức là \(\frac{\left(a+b+c\right)^3}{3abc}-9\geΣ\frac{2c}{a+b}-3\)
tức là \(\frac{\left(a+b+c\right)^3-27abc}{3abc}\geΣ\frac{2c-a-b}{a+b}\)
tức là \(\frac{Σ\left(a^3+3a^2b+3a^2c-7abc\right)}{3abc}\geΣ\frac{c-a-\left(b-c\right)}{a+b}\)
tức là \(\frac{Σ\left(a^3-abc\right)+3Σ\left(a^2b+a^2c-2abc\right)}{3abc}\geΣ\frac{c-a-\left(b-c\right)}{a+b}\)
tức là \(\frac{\frac{1}{2}\left(a+b+c\right)Σ\left(a-b\right)^2+3Σc\left(a-b\right)^2}{3abc}\geΣ\left(a-b\right)\left(\frac{1}{b+c}-\frac{1}{c+a}\right)\)
tức là \(Σ\left(a-b\right)^2\left(\frac{a+b+7c}{6abc}-\frac{1}{\left(a+c\right)\left(b+c\right)}\right)\ge0\)
Đúng theo BĐT AM-GM (giải theo SOS xấu v~)
\(\left(a+b+7c\right)\left(a+c\right)\left(b+c\right)-6abc>\left(a+b\right)\left(a+c\right)\left(b+c\right)-6abc\ge8abc-6abc>0\)
có cho số dương hay j k