Cho a,b, c khác 0.
CMR: \(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\) lớn hơn hoặc bằng \(a+b+c \)
Giải nhanh giúp mk vs nhá
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KK
23 tháng 3 2017
\(\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\)
\(\Leftrightarrow a-\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}+b-\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}+c-\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\)
\(\Leftrightarrow a+b+c-\left[\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}+\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}+\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\right]\)
Áp dụng bất đẳng thức Cauchy - Schwarz cho 3 bộ số thực không âm
\(\Rightarrow\left\{{}\begin{matrix}a^2+ab+b^2\ge3\sqrt[3]{a^3b^3}=3ab\\b^2+bc+c^2\ge3\sqrt[3]{b^3c^3}=3bc\\c^2+ca+a^2\ge3\sqrt[3]{c^3a^3}=3ca\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}\le\dfrac{ab\left(a+b\right)}{3ab}=\dfrac{a+b}{3}\\\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}\le\dfrac{bc\left(b+c\right)}{3bc}=\dfrac{b+c}{3}\\\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\le\dfrac{ca\left(c+a\right)}{3ca}=\dfrac{c+a}{3}\end{matrix}\right.\)
\(\Rightarrow\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}+\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}+\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\le\dfrac{2\left(a+b+c\right)}{3}\)
\(\Leftrightarrow a+b+c-\left[\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}+\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}+\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\right]\ge a+b+c-\dfrac{2\left(a+b+c\right)}{3}\)
\(\Leftrightarrow a+b+c-\left[\dfrac{ab\left(a+b\right)}{a^2+ab+b^2}+\dfrac{bc\left(b+c\right)}{b^2+bc+c^2}+\dfrac{ca\left(c+a\right)}{c^2+ca+a^2}\right]\ge\dfrac{a+b+c}{3}\)
\(\Leftrightarrow\dfrac{a^3}{a^2+ab+b^2}+\dfrac{b^3}{b^2+bc+c^2}+\dfrac{c^3}{c^2+ca+a^2}\ge\dfrac{a+b+c}{3}\) ( đpcm )
Dấu "=" xảy ra khi \(a=b=c\)
11 tháng 2 2020
a,b,c< 0 mà a+b+c bé hơn hoặc bằng 1
a+b+c ít nhất phải bằng 3 chứ!
NC
29 tháng 9 2020
Xét: \(1+c^2=ab+bc+ca+c^2=\left(a+c\right)\left(b+c\right)\)
Tương tự CM được:
\(1+b^2=\left(a+b\right)\left(c+b\right)\) và \(1+a^2=\left(c+a\right)\left(b+a\right)\)
Mặt khác ta tách: \(\hept{\begin{cases}a-b=\left(a+c\right)-\left(b+c\right)\\b-c=\left(a+b\right)-\left(c+a\right)\\c-a=\left(c+b\right)-\left(a+b\right)\end{cases}}\)
Thay vào ta được:
\(Vt=\frac{\left(a+c\right)-\left(b+c\right)}{\left(a+c\right)\left(b+c\right)}+\frac{\left(a+b\right)-\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}+\frac{\left(c+b\right)-\left(a+b\right)}{\left(b+c\right)\left(a+b\right)}\)
\(=\frac{1}{b+c}-\frac{1}{c+a}+\frac{1}{c+a}-\frac{1}{a+b}+\frac{1}{a+b}-\frac{1}{b+c}\)
\(=0\)
=> đpcm
NB
1
30 tháng 1 2019
Áp dụng BĐT Cauchy ta có :
\(\frac{a^3}{bc}+b+c\ge3\sqrt[3]{\frac{a^3bc}{bc}}=3a\)
\(< =>\frac{a^3}{bc}\ge3a-b-c\left(1\right)\)
Chứng minh tương tự => \(\hept{\begin{cases}\frac{b^3}{ca}\ge3b-a-c\left(2\right)\\\frac{c^3}{ab}\ge3c-a-b\left(3\right)\end{cases}}\)
(1),(2),(3) =>\(\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge3a-b-c+3b-a-c+3c-a-b=a+b+c\left(đpcm\right)\)
Dấu "=" xảy ra khi a=b=c
Bạn dùng phương pháp chọn điểm rơi thôi
2 tháng 9 2017
đặt \(\sqrt{\frac{ab}{c}}=x;\sqrt{\frac{bc}{a}}=y;\sqrt{\frac{ca}{b}}=z\Rightarrow xy+yz+zx=1\)
\(P=\frac{ab}{ab+c}+\frac{bc}{bc+a}+\frac{ca}{ca+b}\)
\(=\frac{\frac{ab}{c}}{\frac{ab}{c}+1}+\frac{\frac{bc}{a}}{\frac{bc}{a}+1}+\frac{\frac{ca}{b}}{\frac{ca}{b}+1}=\frac{x^2}{x^2+1}+\frac{y^2}{y^2+1}+\frac{z^2}{z^2+1}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\frac{\left(x+y+z\right)^2}{3}}=\frac{3}{4}\left(Q.E.D\right)\)
N
26 tháng 12 2019
De dung la:
\(\Sigma_{cyc}\frac{1}{1+a^2+b^2}\le\frac{9}{5}\)
\(\Leftrightarrow\Sigma_{cyc}\frac{a^2+b^2}{1+a^2+b^2}\ge\frac{6}{5}\)
\(VT\ge\frac{\left(\Sigma_{cyc}\sqrt{a^2+b^2}\right)^2}{2\Sigma_{cyc}a^2+3}\left(M\right)\)
Consider:
\(VT_M\ge\frac{6}{5}\)
\(5\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge\Sigma_{cyc}a^2+9\)
Consider:
\(5\Sigma_{cyc}\sqrt{\left(a^2+b^2\right)\left(b^2+c^2\right)}\ge5\Sigma_{cyc}a^2+5\Sigma_{cyc}ab=5\Sigma_{cyc}a^2+5\)
Gio can cung minh:
\(5\Sigma_{cyc}a^2+5\ge\Sigma_{cyc}a^2+9\)
\(\Leftrightarrow\Sigma_{cyc}a^2\ge1\)
Ta lai co:
\(\Sigma_{cyc}a^2\ge\Sigma_{cyc}ab=1\)
Dau '=' xay ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)
\(/(/frac//\)la j zay
Áp dụng BĐT Cô si với a,b,c>0 ta có:
\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}.\frac{ca}{b}}=2\sqrt{c^2}=2c\)
Tương tự \(\frac{ca}{b}+\frac{ab}{c}\ge2a\)
\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)
\(\Rightarrow2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\)
\(\Rightarrow\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\ge a+b+c\)