\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\) ( Svac-xơ, Cauchy các kiểu -,- ) 

\(\Leftrightarrow\)\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}}{2}=\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\) ( đpcm ) 

... 

\(2VP=\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

\(\le\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VT\)

Từ đây,ta có: \(2VT\ge2VP\Rightarrow VT\ge VP^{\left(đpcm\right)}\)

1) Áp dụng bđt Cauchy:

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}\ge2\sqrt{\dfrac{1}{a^2b^2}}=\dfrac{2}{ab}\)

Xong

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

\(\Leftrightarrow\dfrac{2}{a+2}-1+\dfrac{2}{b+2}-1+\dfrac{2}{c+2}-1\ge2-3\)

\(\Rightarrow1\ge\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}=\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\)

\(\Rightarrow1\ge\dfrac{\left(a+b+c\right)^2}{a^2+2a+b^2+2b+c^2+2c}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

Phía trên thoả mãn \(\ge1\) chứ không phải 3/2 đâu ạ 

Cách 1: Áp dụng BĐT Cauchy

\(\dfrac{a}{b^2}+\dfrac{1}{a}\ge2\sqrt{\dfrac{a}{b^2}.\dfrac{1}{a}}=\dfrac{2}{b}\)

Tương tự: \(\dfrac{b}{c^2}+\dfrac{1}{b}\ge\dfrac{2}{c}\)

\(\dfrac{c}{a^2}+\dfrac{1}{c}\ge\dfrac{2}{a}\)

Cộng vế theo vế các BĐT vừa chứng minh rồi rút gọn, ta có:

\(\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)(đpcm)

Đẳng thức xảy ra khi \(a=b=c\)

Cách 2: Áp dụng BĐT Bunyakovsky

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

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

\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le\dfrac{a}{b^2}+\dfrac{b}{c^2}+\dfrac{c}{a^2}\)(đpcm)

Đẳng thức xảy ra khi \(a=b=c\)

3/ Áp dụng bất đẳng thức AM-GM, ta có :

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)

\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)

\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)

Cộng 3 vế của BĐT trên ta có :

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

\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)

Bài 1:

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)

Do đó:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)

Dấu "=" xảy ra khi $a=b=c$

Bài 2:

Thay $1=a+b+c$ và áp dụng BĐT AM-GM ta có:

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

\(=\frac{(a+a+b+c)(b+a+b+c)(c+a+b+c)}{abc}\)

\(\geq \frac{4\sqrt[4]{a.a.b.c}.4\sqrt[4]{b.a.b.c}.4\sqrt[4]{c.a.b.c}}{abc}=\frac{64abc}{abc}=64\)

Ta có đpcm
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$