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1)\(\dfrac{c-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}+\dfrac{a-c}{\left(b-a\right)\left(b-c\right)\left(a-c\right)}+\dfrac{b-a}{\left(b-a\right)\left(c-b\right)\left(c-a\right)}=\dfrac{c-b+a-c+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
Lời giải:
\(\text{VT}=\frac{1}{a(a-b)(a-c)}+\frac{1}{b(b-c)(b-a)}+\frac{1}{c(c-a)(c-b)}\)
\(=\frac{bc(c-b)}{abc(a-b)(b-c)(c-a)}+\frac{ac(a-c)}{abc(a-b)(b-c)(c-a)}+\frac{ab(b-a)}{abc(a-b)(b-c)(c-a)}\)
\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{abc(a-b)(b-c)(c-a)}\) (1)
Xét \(bc(c-b)+ac(a-c)+ab(b-a)=bc(c-b)-ac[(c-b)+(b-a)]+ab(b-a)\)
\(=(c-b)(bc-ac)+(b-a)(ab-ac)=c(c-b)(b-a)+a(b-a)(b-c)\)
\(=(c-b)(b-a)(c-a)=(a-b)(b-c)(c-a)\) (2)
Từ \((1),(2)\Rightarrow \text{VT}=\frac{(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}=\frac{1}{abc}\)
Ta có đpcm.
B1:
\(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Xét hiệu:
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\)
\(=\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)
=> BĐT luôn đúng
*
Ta có:
\(a< b+c\Rightarrow a^2< ab+ac\)
\(b< a+c\Rightarrow b^2< ab+ac\)
\(c< a+b\Rightarrow a^2< ac+bc\)
Cộng từng vế bất đẳng thức ta được:
\(a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
Vậy: \(ab+bc+ca\le a^2+b^2+c^2< 2\left(ab+bc+ca\right)\)
B2:
Ta có: \(a+b>c\) ; \(b+c>a\); \(a+c>b\)
Xét:\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+b+a+b}=\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{a+b+c}+\dfrac{1}{a+c+b}=\dfrac{2}{a+b+c}>\dfrac{2}{b+c+b+c}=\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+b+c}+\dfrac{1}{b+c+a}=\dfrac{2}{a+b+c}>\dfrac{2}{a+c+a+c}=\dfrac{1}{a+c}\)
Suy ra:
\(\dfrac{1}{a+c}+\dfrac{1}{b+c}>\dfrac{1}{a+b}\)
\(\dfrac{1}{a+b}+\dfrac{1}{a+c}>\dfrac{1}{b+c}\)
\(\dfrac{1}{a+b}+\dfrac{1}{b+c}>\dfrac{1}{a+c}\)
=> ĐPCM
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)=8\)
\(\Leftrightarrow\dfrac{a+b}{a}\times\dfrac{b+c}{b}\times\dfrac{a+c}{c}=8\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\)
~*~*~*~*~
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(=\dfrac{3}{4}+\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\) (1)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{ab}{\left(a+b\right)\left(b+c\right)}+\dfrac{b}{b+c}-\dfrac{bc}{\left(b+c\right)\left(c+a\right)}+\dfrac{c}{c+a}-\dfrac{ac}{\left(c+a\right)\left(a+b\right)}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\left(1-\dfrac{b}{b+c}\right)+\dfrac{b}{b+c}\left(1-\dfrac{c}{c+a}\right)+\dfrac{c}{a+c}\left(1-\dfrac{a}{a+b}\right)\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{a}{a+b}\times\dfrac{c}{b+c}+\dfrac{b}{b+c}\times\dfrac{a}{a+c}+\dfrac{c}{a+c}\times\dfrac{b}{a+b}\)
\(=\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)}{\left(a+c\right)\left(b+c\right)\left(a+b\right)}=\dfrac{3}{4}\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)=\dfrac{3}{4}\times8abc\)
\(\Leftrightarrow ac\left(a+c\right)+ab\left(a+b\right)+bc\left(b+c\right)+2abc=8abc\)
\(\Leftrightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=8abc\) luôn đúng
=> (1) đúng
Bạn cũng có thể giải bằng cách đặt \(x=\dfrac{a}{a+b};y=\dfrac{b}{b+c};z=\dfrac{c}{a+c}\).
\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ca\right)\left(a-abc\right)\)
\(\Leftrightarrow a^2b+ab^2c^2-a^3bc-b^2c=b^2a+a^2bc^2-ca^2-ab^3c\)
\(\Leftrightarrow a^2b-ab^2-b^2c+ca^2=a^2bc^2-ab^3c+a^3bc-ab^2c^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=abc\left(a-b\right)\left(a+b+c\right)\)
\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)