Cho a, b,c >0 và a+b=c = 3.Cm : \(\dfrac{1}{2+a^2b}\)+\(\dfrac{1}{2+b^2c}\)+\(\dfrac{1}{2+c^2a}\)\(\ge\)1
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\(VT=\dfrac{a^2}{b+ab^2c}+\dfrac{b^2}{b+abc^2}+\dfrac{c^2}{c+a^2bc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}=\dfrac{9}{3+3abc}\)
\(VT\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^3}{9}}=\dfrac{3}{2}\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
Lời giải:
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\frac{\left ( \frac{a}{bc} \right )^2}{\frac{1}{c}}+\frac{\left ( \frac{b}{ca} \right )^2}{\frac{1}{a}}+\frac{\left ( \frac{c}{ab} \right )^2}{\frac{1}{b}}\geq \frac{\left ( \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
\(\Leftrightarrow \text{VT}\geq \frac{\left ( \frac{a^2+b^2+c^2}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\)
Theo hệ quả của BĐT AM-GM thì:
\(a^2+b^2+c^2\geq ab+bc+ac\)
\(\Rightarrow \text{VT}\geq \frac{\left ( \frac{ab+bc+ac}{abc} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right )^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
Ta có đpcm.
Dấu bằng xảy ra khi \(a=b=c\)
Áp dụngk BĐt cô-si, ta có
\(\frac{a^2}{b^2c}+\frac{b^2}{c^2a}+\frac{1}{a}\ge3.\frac{1}{c}\)
Tương tự , rồi cộng vào, ta có
\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)
^_^
\(P=\dfrac{a^2}{ab+\dfrac{1}{b}}+\dfrac{b^2}{bc+\dfrac{1}{c}}+\dfrac{c^2}{ca+\dfrac{1}{a}}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}\)
\(P\ge\dfrac{3\left(ab+bc+ca\right)}{ab+bc+ca+\dfrac{ab+bc+ca}{abc}}=\dfrac{3}{1+\dfrac{1}{abc}}=\dfrac{3abc}{1+abc}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Với a, b, c > 0 có:
\(P=\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\\ =\dfrac{a^2}{a\left(b+2c\right)}+\dfrac{b^2}{b\left(c+2a\right)}+\dfrac{c^2}{c\left(a+2b\right)}\)
\(\Rightarrow P\ge\dfrac{\left(a+b+c\right)^2}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{\left(1+\alpha\right)\left(ab+bc+ca\right)}\)
chọn \(\alpha=\dfrac{1}{abc}\Rightarrow dpcm\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
=> bc+ac+ab=0
ta có
\(bc+ac=-ab\)
<=> \(\left(bc+ac\right)^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)
<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)
tương tự
\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)
\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)
thay vào E ta đc
\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)
=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)
Hình như sai đề :
Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)
\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)
\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )
Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)
\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)
\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )
CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )
Thay ( * ) và ( * ') vào E , ta được :
\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)
\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)
\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)
\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)
\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)
\(=\dfrac{0}{2}=0\)
Vậy \(E=0\)