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\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}=3+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)Ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2;\dfrac{c}{a}+\dfrac{a}{c}\ge2;\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3+2+2+2=9\)
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...
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Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)
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)
\(\Rightarrow\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Có \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
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
a) theo định lý côsi :
\(\dfrac{a}{b}\)+\(\dfrac{b}{a}\)luôn >=2 với mọi a, b , a.b > 0
\(a,\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)\(\Leftrightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge3+2+2+2=9\Rightarrowđpcm\)b, Đặt \(x=b+c;y=a+c;a+b=z\)
Khi đó :
\(=\dfrac{1}{2}\left[\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\left(\dfrac{z}{x}+\dfrac{x}{z}\right)+\left(\dfrac{z}{y}+\dfrac{y}{z}\right)-3\right]\) \(\ge\dfrac{1}{2}\left(2+2+2-3\right)=1,5\Rightarrowđpcm\)
Đầu tiên ta cm:\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)(tự cm)
Áp dụng:\(\Rightarrow\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ac}+\dfrac{1}{c^2+2ab}\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}\)
Lại có:\(a^2+b^2+c^2+2ab+2bc+2ca=\left(a+b+c\right)^2\le1\)
\(\Rightarrow\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}\ge\dfrac{9}{1}=9\)
\(\Rightarrowđpcm\)
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 \(a=b=c=\dfrac{1}{3}\)
dạng này chắc chắc là phải dùng AM-GM ngược dấu rồi :)
Ta có:
\(\dfrac{1+b}{1+4a^2}=1+b-\dfrac{4a^2\left(b+1\right)}{4a^2+1}\ge1+b-\dfrac{4a^2\left(b+1\right)}{4a}=1+b-a\left(b+1\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\dfrac{1+c}{1+4b^2}\ge1+c-b\left(c+1\right);\dfrac{1+a}{1+4c^2}\ge1+a-c\left(a+1\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(VT=\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+c^2}\)
\(\ge3+\left(a+b+c\right)-\left(ab+bc+ca\right)-\left(a+b+c\right)\)
\(=3-\dfrac{1}{3}\left(a+b+c\right)^2=3-\dfrac{1}{3}\cdot\dfrac{9}{4}=\dfrac{9}{4}=VP\)
Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{2}\)
\(VT=\left(\dfrac{a}{1+4c^2}+\dfrac{b}{1+4a^2}+\dfrac{c}{1+4b^2}\right)+\left(\dfrac{1}{1+4c^2}+\dfrac{1}{1+4a^2}+\dfrac{1}{1+4b^2}\right)\)
\(VT=\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)+3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Xét \(\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2a}{1+4c^2}\le\dfrac{4c^2a}{4c}=ca\\\dfrac{4a^2b}{1+4a^2}\le\dfrac{4a^2b}{4a}=ab\\\dfrac{4b^2c}{1+4b^2}\le\dfrac{4b^2c}{4b}=bc\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\ge\dfrac{3}{2}-\left(ab+bc+ca\right)\) (1)
Xét \(3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2}{1+4c^2}\le\dfrac{4c^2}{4c}=c\\\dfrac{4a^2}{1+4a^2}\le\dfrac{4a^2}{4a}=a\\\dfrac{4b^2}{1+4b^2}\le\dfrac{4b^2}{4b}=b\end{matrix}\right.\)
\(\Rightarrow3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\ge\dfrac{3}{2}\) (2)
Từ (1) và (2)
\(\Rightarrow VT\ge\dfrac{3}{2}-\left(ab+bc+ca\right)+\dfrac{3}{2}\)
\(\Rightarrow VT\ge3-\left(ab+bc+ca\right)\) (3)
Theo hệ quả của bất đẳng thức Cauchy
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{3}{4}\ge ab+bc+ca\)
\(\Rightarrow3-\dfrac{3}{4}\le3-\left(ab+bc+ca\right)\)
\(\Rightarrow\dfrac{9}{4}\le3-\left(ab+bc+ca\right)\) (4)
Từ (3) và (4)
\(\Rightarrow VT\ge\dfrac{9}{4}\)
\(\Leftrightarrow\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+4c^2}\ge\dfrac{9}{4}\) (đpcm)
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{2}\)
-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 \(a=b=c=\dfrac{1}{3}\)