CM NẾU \(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}=\frac{1}{ABC}\);\(\cdot\left(ABC\ne0\right)\)VÀ\(A+B+C\ne0\)THÌ \(\frac{1}{A^n}+\frac{1}{B^n}+\frac{1}{C^n}=\frac{1}{A^n+B^n+C^n}\)
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HP
20 tháng 11 2016
\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)
\(< =>\frac{a^2}{b+c}+a+\frac{b^2}{a+c}+b+\frac{c^2}{a+b}+c=a+b+c\)
\(< =>\frac{a^2+a\left(b+c\right)}{b+c}+\frac{b^2+b\left(c+a\right)}{c+a}+\frac{c^2+c\left(a+b\right)}{a+b}=a+b+c\)
\(< =>\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=1\) (chia cả 2 vế cho a+b+c)
AH
Akai Haruma
Giáo viên
31 tháng 7 2019
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)
\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$
13 tháng 1 2020
Quy đồng full :)
\(\frac{1}{a\left(1+b\right)}+\frac{1}{b\left(1+c\right)}+\frac{1}{c\left(1+a\right)}\ge\frac{3}{1+abc}\)
\(\Leftrightarrow\frac{1+abc}{a\left(1+b\right)}+\frac{1+abc}{b\left(1+c\right)}+\frac{1+abc}{c\left(1+a\right)}\ge3\)
\(\Leftrightarrow\left[\frac{1+abc}{a\left(1+b\right)}+1\right]+\left[\frac{1+abc}{b\left(1+c\right)}+1\right]+\left[\frac{1+abc}{c\left(1+a\right)}+1\right]\ge6\)
\(\Leftrightarrow\frac{1+abc+ab+a}{a\left(1+b\right)}+\frac{1+abc+bc+b}{b\left(1+c\right)}+\frac{1+abc+c+ac}{c\left(1+a\right)}\ge6\)
\(\Leftrightarrow\frac{ab\left(c+1\right)+\left(a+1\right)}{a\left(1+b\right)}+\frac{bc\left(a+1\right)+\left(b+1\right)}{b\left(1+c\right)}+\frac{ac\left(b+1\right)+\left(c+1\right)}{c\left(1+a\right)}\ge6\)
\(\Leftrightarrow\frac{b\left(c+1\right)}{1+b}+\frac{a+1}{a\left(1+b\right)}+\frac{c\left(a+1\right)}{1+c}+\frac{b+1}{b\left(1+c\right)}+\frac{a\left(b+1\right)}{1+a}+\frac{c+1}{c\left(1+a\right)}\ge6\)
Ta có vế trái tương đương với:
\(\left[\frac{b\left(c+1\right)}{1+b}+\frac{b+1}{b\left(c+1\right)}\right]+\left[\frac{a\left(b+1\right)}{1+a}+\frac{1+a}{a\left(b+1\right)}\right]+\left[\frac{c\left(a+1\right)}{1+c}+\frac{1+c}{c\left(a+1\right)}\right]\)
\(\ge2+2+2=6\)
=> đpcm
vô đây mà xem ; /hoi-dap/question/125436.html?pos=554506