BDT

\(x+\dfrac{1}{x}=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2+2\ge2\)

nhân PP vào là ra

\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3+2+2+2=9\)

Theo BĐT Cauchy:

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

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

\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c\)

Áp dụng BĐT C-S, ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

\(\Rightarrow VT\ge9\)

Đẳng thức xảy ra khi a=b=c

Áp dụng BĐT AM - GM, ta có:

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

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

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

\(\ge3+2+2+2=9\)

Dấu "=" xảy ra khi a = b = c

Áp dụng bất đẳng thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\) có:

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

Dấu " = " khi a = b = c

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)

b)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+b+c}\)

=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)

Giải:

Áp dụng BĐT Cô si cho 3 số dương ta được:

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

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

Nhân theo vế 2 BĐT trên ta có:

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

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Nếu đề là \(x,y,z\ge0\) thì làm như sau:

Áp dụng bất đẳng thức Cauchy ta có:

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

Đẳng thức xảy ra khi a = b = c

\(\Rightarrowđpcm\)

Nếu đề là \(a,b,c\ge0\) thì làm như sau:

Áp dụng bất đẳng thức Cauchy ta có:

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

Đẳng thức xảy ra khi a = b = c

\(\Rightarrowđpcm\)

a,b,c là các số dương nên \(\left(a+b+c\right)>=3\cdot\sqrt[3]{abc}\)

\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\)

Do đó: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)

a)Theo bất đẳng thức cauchy:

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge\dfrac{4}{a+b}.\left(a+b\right)\)

\(\Rightarrow\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

Dấu "=" xảy ra khi: \(a=b\)

Ta có điều phải chứng minh

b)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(a+b+c\right)\ge\dfrac{9}{a+b+c}.\left(a+b+c\right)\)

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

Dấu "=" xảy ra khi:

\(a=b=c\)

Ta có điều phải chứng minh

\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)

Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)

\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{x}{2}+\dfrac{x}{2}+\dfrac{1}{16x^2}\right)+\left(\dfrac{y}{2}+\dfrac{y}{2}+\dfrac{1}{16y^2}\right)+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)

\(P\ge3\sqrt[3]{\dfrac{x^2}{64x^2}}+3\sqrt[3]{\dfrac{y^2}{64y^2}}+\dfrac{15}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)

\(P\ge\dfrac{3}{2}+\dfrac{15}{32}\left(\dfrac{4}{x+y}\right)^2\ge\dfrac{3}{2}+\dfrac{15}{32}.\left(\dfrac{4}{1}\right)^2=9\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)

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

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

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

Áp dụng BĐT Cô - si cho 2 số không âm:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\)

\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)

Suy ra:

\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\ge2+2+2+3=9\)

Dấu "=" xảy ra khi: a = b = c

Áp dụng BĐT Cauchy dạng Engel , ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(\dfrac{9}{a+b+c}\)

⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\left(a+b+c\right).\dfrac{9}{a+b+c}\)

⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}9\)

\("="\text{⇔}a=b=c\)

Áp dụng bất đẳng thức Cô - si cho 3 số không âm ta có :

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

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

Nhân vế theo vế ta có :

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

Dấu " = " xảy ra khi a = b = c .