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30 tháng 4 2017

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}=3+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)Ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2;\dfrac{c}{a}+\dfrac{a}{c}\ge2;\dfrac{b}{c}+\dfrac{c}{b}\ge2\)

\(\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3+2+2+2=9\)

30 tháng 4 2017

Gọi \(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) Ta có:

\(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)\(=\left(\dfrac{a}{a}+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{b}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{c}{c}\right)\)

\(=\left(1+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+1+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+1\right)\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)

\(=3+\left(\dfrac{a^2}{ab}+\dfrac{b^2}{ab}\right)+\left(\dfrac{b^2}{bc}+\dfrac{c^2}{bc}\right)+\left(\dfrac{c^2}{ac}+\dfrac{a^2}{ac}\right)\)

\(=3+\dfrac{a^2+b^2}{ab}+\dfrac{b^2+c^2}{bc}+\dfrac{c^2+a^2}{ac}\)

\(=3+\dfrac{a^2-2ab+b^2+2ab}{ab}+\dfrac{b^2-2bc+c^2+2bc}{bc}+\dfrac{c^2-2ac+a^2+2ac}{ac}\)

\(=3+\dfrac{\left(a-b\right)^2+2ab}{ab}+\dfrac{\left(b-c\right)^2+2bc}{bc}+\dfrac{\left(c-a\right)^2+2ac}{ac}\)

\(=3+\dfrac{\left(a-b\right)^2}{ab}+2+\dfrac{\left(b-c\right)^2}{bc}+2+\dfrac{\left(c-a\right)^2}{ac}+2\)

\(=9+\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ac}\)

Ta thấy: \(\dfrac{\left(a-b\right)^2}{ab}\ge0\) với \(\forall\) a, b

\(\dfrac{\left(b-c\right)^2}{bc}\ge0\) với \(\forall\) b, c

\(\dfrac{\left(c-a\right)^2}{ac}\ge\) 0 với \(\forall\) a, c

=> \(A\ge9\).

Vậy...

Chúc bạn học tốt!

12 tháng 5 2022

-Áp dụng BĐT Caushy Schwarz ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)

-Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

1 tháng 1 2018

Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)

8 tháng 4 2018

\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Áp dụng BĐT Cô-si, ta có:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}\ge2\)

C/m tương tự, ta có:

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)

\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2+2+2+3\)

\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge9\)

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)

áp dụng BĐT:
1/a +1/b+1/c>= 9/a+b+c mà a+b+c=1

=>1/a+1/b+1/c≥9

9 tháng 4 2018

a)\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

=\(\dfrac{a}{a}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{c}+\dfrac{c}{a}+\dfrac{c}{b}\)

=\(1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)

=3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\)

áp dụng BĐT cô si ta có

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\)

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)

cmtt ta có \(\dfrac{b}{c}+\dfrac{c}{b}\ge2\); \(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)

=> 3+\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)\ge9\)

=> \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)

a)Áp dụng bđt AM-GM cho 3 số không âm ta có:

\(a+b+c\ge3\sqrt[3]{abc}\)

TT\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\)

Nhân vế theo vế ta có:\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\dfrac{1}{abc}}=9\left(đpcm\right)\)

b)\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)

\(=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\)

Svac-xo:

\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)

Lại có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(tự cm)

\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)

\(\Rightarrowđpcm\)

7 tháng 2 2021

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19 tháng 4 2018

a) theo định lý côsi :

\(\dfrac{a}{b}\)+\(\dfrac{b}{a}\)luôn >=2 với mọi a, b , a.b > 0

19 tháng 6 2017

\(a,\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge3+2+2+2=9\Rightarrowđpcm\)b, Đặt \(x=b+c;y=a+c;a+b=z\)

Khi đó :

\(=\dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)-3\right]\) \(\ge\dfrac{1}{2}\left(2+2+2-3\right)=1,5\Rightarrowđpcm\)

19 tháng 6 2017

Camon nha~

9 tháng 4 2017

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+2bc+b^2+2ac+c^2+2ab}\)

\(=\dfrac{3^2}{\left(a+b+c\right)^2}=\dfrac{9}{\left(a+b+c\right)^2}=9\left(a+b+c\le1\right)\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)