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29 tháng 11 2016
\(\frac{a^3}{b+c}+\frac{b^3}{a+c}+\frac{c^3}{a+b}=\frac{a^4}{ab+ac}+\frac{b^4}{ba+bc}+\frac{c^4}{ca+cb}\)
\(\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ca\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{1}{2}\)
20 tháng 3 2020
BĐT phụ:\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x-y\right)^2\ge0\left(true\right)\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{4}{a+b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) ( đpcm )
Vậy.......
17 tháng 10 2016
Từ giả thiết ta có \(1+c^2=ab+bc+ac+c^2=\left(a+c\right)\left(b+c\right)\) ; \(1+a^2=ab+bc+ac+a^2=\left(a+b\right)\left(a+c\right)\)
\(1+b^2=ab+bc+ac+b^2=\left(b+a\right)\left(b+c\right)\)
Suy ra \(\frac{a+b}{1+c^2}+\frac{b+c}{1+a^2}+\frac{c+a}{1+b^2}=\frac{a+b}{\left(c+a\right)\left(c+b\right)}+\frac{b+c}{\left(a+b\right)\left(a+c\right)}+\frac{c+a}{\left(b+a\right)\left(b+c\right)}\)
\(=\frac{\left(a+b\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}+\frac{\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Theo BĐT Cauchy , ta có : \(\frac{\left(a+b\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(a+b\right)^2}{\left(a+b+b+c+c+a\right)^3}=\frac{27\left(a+b\right)^2}{8\left(a+b+c\right)^3}\)
Tương tự : \(\frac{\left(b+c\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(b+c\right)^2}{8\left(a+b+c\right)^3}\) ; \(\frac{\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{27\left(c+a\right)^2}{8\left(a+b+c\right)^3}\)
\(\Rightarrow\frac{\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge\frac{9}{8\left(a+b+c\right)^3}.3\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]\)
\(\ge\frac{9}{8\left(a+b+c\right)^3}.\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^2\) (Áp dụng BĐT Bunhiacopxki)
\(=\frac{9.4\left(a+b+c\right)^2}{8\left(a+b+c\right)^3}=\frac{9}{2\left(a+b+c\right)}\) (đpcm)
A
0
22 tháng 3 2019
Áp dụng bất đẳng thức Cô-si ta có:
\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)
Cộng theo vế ta được:
\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}\ge\frac{9}{a+b+c}\)
\(\Leftrightarrow\left(ab+ac+bc\right)\left(a+b+c\right)-9abc\ge0\)
\(\Leftrightarrow a^2b+a^2c+abc+abc+ab^2+b^2c+abc+ac^2+bc^2-9abc\ge0\)
\(\Leftrightarrow a^2b+a^2c+ab^2+b^2c+ac^2+bc^2-6abc\ge0\)
\(\Leftrightarrow\left(a^2b-2abc+bc^2\right)+\left(a^2c-2abc+b^2c\right)+\left(ab^2-2abc+ac^2\right)\ge0\)
\(\Leftrightarrow b\left(a-b\right)^2+c\left(a-c\right)^2+a\left(b-c\right)^2\ge0\)(luôn đúng \(\forall a;b;c>0\))
Vật bđt đã đc chứng minh
Cho a,b,c>0 thì dễ thôi :v
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}\)
Khi a=b=c