Cho a,b,c>0.Cmr
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\)
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3 tháng 1 2019
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 1 2019
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
22 tháng 7 2018
Ta có BĐT : \(\dfrac{1}{a}+\dfrac{1}{b}\) ≥ \(\dfrac{4}{a+b}\) ( \(a,b>0\) )
\(\dfrac{1}{b}+\dfrac{1}{c}\text{≥}\dfrac{4}{b+c}\left(b;c>0\right)\)
\(\dfrac{1}{a}+\dfrac{1}{c}\text{≥}\dfrac{4}{a+c}\left(a;c>0\right)\)
Cộng từng vế của các BĐT trên , ta có :
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{a+c}\)
⇔ \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\text{≥}\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{a+c}\)
22 tháng 7 2018
Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế theo vế ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge2\left(\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\right)\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
\(\Rightarrowđpcm\)
23 tháng 7 2018
Đặt \(A=\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{a+c}{b}\)
\(=\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}\)
Áp dụng bất đẳng thức :
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\dfrac{b}{a}+\dfrac{a}{b}\ge2\)
\(\Rightarrow\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{b}{a}+\dfrac{a}{b}\ge6\)
Đặt \(B=\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{a+c}\)
\(\Rightarrow B+3=\dfrac{c}{a+b}+1+\dfrac{a}{b+c}+1+\dfrac{b}{a+c}+1\)
\(=\dfrac{a+b+c}{a+b}+\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{c+a}\)
\(=\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
Ta có : \(2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(=\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\)
\(\Rightarrow B+3\ge\dfrac{9}{2}\Rightarrow B\ge\dfrac{3}{2}\)
\(\Rightarrow A+B\ge\dfrac{15}{2}\)
Dấu " = " xảy ra khi a = b = c .
AH
Akai Haruma
Giáo viên
28 tháng 11 2018
Lời giải:
Áp dụng BĐT Cauchy-Schwarz và AM-GM ta có:
\(\text{VT}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+abc(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+(ab+bc+ac)+\frac{a^2}{ab}+\frac{b^2}{bc}+\frac{c^2}{ac}\)
\(\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+(ab+bc+ac)+\frac{(a+b+c)^2}{ab+bc+ac}\)
\(\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2\sqrt{(ab+bc+ac).\frac{(a+b+c)^2}{ab+bc+ac}}\)
\(=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+2(a+b+c)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c+(a+b+c)\)
\(\geq 6\sqrt[6]{\frac{1}{a}.\frac{1}{b}.\frac{1}{c}.a.b.c}+(a+b+c)=6+a+b+c\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=1$
AH
Akai Haruma
Giáo viên
29 tháng 1 2018
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 \(a=b=c\)
S
3 tháng 3 2019
\(vì:a,b,c>0\Rightarrow\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}>0\)
\(Cosi:\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\ge\dfrac{2}{\dfrac{a+b}{2}}=\dfrac{4}{a+b}\)
\(\dfrac{4}{2a+b+c}\le\dfrac{1}{4}\left(\dfrac{4}{a+b}+\dfrac{4}{a+c}\right)\le\dfrac{1}{16}\left(\dfrac{8}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}.tươngtự:\dfrac{4}{a+b+2c}\le\dfrac{1}{4a}+\dfrac{1}{4b}+\dfrac{1}{2c};\dfrac{4}{a+2b+c}\le\dfrac{1}{4a}+\dfrac{1}{2b}+\dfrac{1}{2c}.\text{cộng vế theo vế ta được:}\dfrac{4}{a+2b+c}+\dfrac{4}{2a+b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(\text{đpcm}\right)\)
NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 3 2019
Áp dụng BĐT \(\dfrac{1}{x+y+z+t}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{t}\right)\) với các số dương
Ta có: \(\dfrac{4}{a+a+b+c}\le\dfrac{4}{16}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\dfrac{4}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)
\(\dfrac{4}{a+2b+c}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)
Cộng vế với vế:
\(\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\le\dfrac{1}{4}\left(\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
Dấu "=" xảy ra khi \(a=b=c\)
VT
28 tháng 1 2018
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
\(2A+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow A\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\left(ĐPCM\right)\)
^_^
AT
1
AH
Akai Haruma
Giáo viên
27 tháng 9 2018
Lời giải:
Từ \(a+b+c\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow a+b+c\geq \frac{ab+bc+ac}{abc}\Rightarrow abc(a+b+c)\geq ab+bc+ac\)
\(\Rightarrow a^2b^2c^2(a+b+c)^2\geq (ab+bc+ac)^2(1)\)
Áp dụng BĐT AM-GM:
\(a^2b^2+b^2c^2\geq 2ab^2c\)
\(b^2c^2+c^2a^2\geq 2abc^2\)
\(a^2b^2+c^2a^2\geq 2a^2bc\)
Cộng theo vế, rút gọn \(\Rightarrow a^2b^2+b^2c^2+c^2a^2\geq abc(a+b+c)\)
\(\Rightarrow (ab+bc+ac)^2\geq 3abc(a+b+c)(2)\)
Từ \((1);(2)\Rightarrow a^2b^2c^2(a+b+c)^2\geq 3abc(a+b+c)\)
\(\Rightarrow abc(a+b+c)\geq 3\Rightarrow a+b+c\geq \frac{3}{abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Ta có:
\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+1=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{b^2}{b^2}\)
\(\ge\dfrac{\left(a+2b+c\right)^2}{ab+bc+ca+b^2}=\dfrac{\left(a+b\right)^2+2\left(a+b\right)\left(b+c\right)+\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)}\)
\(=\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+2\)
Sorry bác Neet tới đây e bí mất
Ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge3.\sqrt[3]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}}=3\)(1)
\(\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}\ge2.\sqrt{\dfrac{a+b}{b+c}.\dfrac{b+c}{a+b}}=2\)
\(\Leftrightarrow\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\ge3\)(2)
Từ (1), (2), ta có: \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-\dfrac{a+b}{b+c}-\dfrac{b+c}{a+b}-1\ge0\)
\(\Leftrightarrow\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge\dfrac{a+b}{b+c}+\dfrac{b+c}{a+b}+1\)
Dấu "=" xảy ra khi \(a=b=c\)