c) Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có :

\(\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}\ge\dfrac{\left(1+1+1\right)^2}{A+B+C}=\dfrac{9}{A+B+C}\)

Dấu "=" xảy ra khi và chỉ khi\(\dfrac{1}{A}=\dfrac{1}{B}=\dfrac{1}{C}\)

a,

(a+ b)(\(\frac{1}{a}\)+\(\frac{1}{b}\)) =1+\(\frac{a}{b}\)+\(\frac{b}{a}\)+1 =2+\(\frac{a}{b}\)+\(\frac{b}{a}\)>=4    {vì\(\frac{a}{b}\)+\(\frac{b}{a}\)>=2 theo bất đẳng thức cô-si }.dau"="xay ra khi va  chi khi a=b

b,

(a+b+c)(1/a+1/b+1/c)=1+a/b+a/c+1+b/a+b/c+1+c/a+c/b

=3+(\(\frac{a}{b}\)+\(\frac{b}{a}\))+(\(\frac{b}{c}\)+\(\frac{c}{b}\))+(\(\frac{a}{c}\)+c/a)>=3+2+2+2=9

đầu"="xảy ra khi và chỉ khi a=b=c                                 {>= có nghĩa là lớn hơn hoặc bằng}

* Áp dụng BĐT \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với $x,y>0$ vào bài toán có :

\(\dfrac{1}{4}\cdot\left(\dfrac{4}{a+b}\right)\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\)

Cộng vế với vế các BĐT có :

\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

Áp dụng bất đẳng thức Bunyakovsky ta có:

\(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\right)\)

\(=\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left[\left(\dfrac{1}{\sqrt{a}}\right)^2+\left(\dfrac{2}{\sqrt{b}}\right)^2+\left(\dfrac{3}{\sqrt{c}}\right)^2\right]\)

\(\ge\left(\sqrt{a}.\dfrac{1}{\sqrt{a}}+\sqrt{b}.\dfrac{2}{\sqrt{b}}+\sqrt{c}.\dfrac{3}{\sqrt{c}}\right)^2\)

\(=\left(1+2+3\right)^2=36\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}a=\dfrac{1}{6}\\b=\dfrac{1}{3}\\c=\dfrac{1}{2}\end{matrix}\right.\)

Vì đề không cho a\(\ne\) b\(\ne\)c

Theo đề ta có:

\(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\)=> \(\dfrac{1+4+9}{a+b+c}=\dfrac{14}{1}=14\)

Mà 14> 36

Nên \(\dfrac{1}{a}+\dfrac{4}{b}+\dfrac{9}{c}\) =14 > 36 ( đpcm)

------ nếu sai thì sory -------------

bn ơi mk cx thích CONAN .......lắm luôn ớ

kết bn nha

Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:

\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).

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

Cộng vế với vế các BĐT trên ta được:

\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge4\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\)

Dấu "=" khi a=b=c

b) \(\dfrac{1}{3a+2b+c}\le\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le\dfrac{1}{36}\left(\dfrac{3}{a}+\dfrac{2}{b}+\dfrac{1}{c}\right)\)

Tương tự cho 2 cái kia rồi cộng lại

\(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{6}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}.16=\dfrac{8}{3}\)

Đẳng thức xảy ra \(\Leftrightarrow\) ... \(\Leftrightarrow a=b=c=\dfrac{3}{16}\)

\(\dfrac{1}{a^2+a+1}\ge\dfrac{1}{a^2+\dfrac{a^2+1}{2}+1}=\dfrac{2}{3}.\dfrac{1}{a^2+1}=\dfrac{2}{3}\left(1-\dfrac{a^2}{a^2+1}\right)\ge\dfrac{2}{3}\left(1-\dfrac{a}{2}\right)\)

Tương tự và cộng lại: \(VT\ge\dfrac{2}{3}\left(3-\dfrac{a+b+c}{2}\right)=\dfrac{2}{3}.\dfrac{3}{2}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)