Lời giải:
Do $a,b,c\in [0;1]$ nên:

$a^2(1-b)\leq 0$

$b^2(1-c)\leq 0$

$c^2(1-a)\leq 0$

Cộng theo vế suy ra: $a^2+b^2+c^2\leq a^2b+b^2c+c^2a$ 

Ta có đpcm.

Áp dụng BĐT Bunyakovski\(,\) ta có: \(\left(a^2b+b^2c+c^2a\right)\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{a}\right)\ge\left(a+b+c\right)^2\)

Do đó: \(VT\ge\frac{\left(a+b+c\right)^3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{abc\left(a+b+c\right)^3}{ab+bc+ca}\ge9abc\)

Bất đẳng thức cuối tương đương: \(\left(a+b+c\right)^3\ge9\left(ab+bc+ca\right)\) \((\ast)\)

Có: \(3=a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\)

\(\therefore\left(ab+bc+ca\right)=\frac{\left(a+b+c\right)^2-3}{2}\)

\((\ast)\) \(\Leftrightarrow\left(a+b+c\right)^3\ge\frac{9}{2}\)\(\Big[(a+b+c)^2-3\Big] \)

\(\Leftrightarrow\frac{1}{2}\left(2a+2b+2c+3\right)\left(a+b+c-3\right)^2\ge0\)

Bất đẳng thức cuối hiển nhiên.

Đẳng thức xảy ra khi \(a=b=c=1\). Done.

Không muốn cách dễ hiểu như trên thì dùng cách khó hiểu một tí cũng hong sao :3

Giả sử \(c=\min\{a,b,c\}\)\(,\) ta có:

\(\text{VT-VP}={\frac { \left( a+b+c \right) \Big[{c}^{2} \left( a-b \right) ^{2} \left( a+b \right) +{a}^{2} \left( b-c \right) \left( b+c \right) \left( a- c \right) \Big]}{ab+ac+bc}}+{\frac {abc \left( 2\,a+2\,b+2\,c+3 \right) \left( a+b+c-3 \right) ^{2}}{2\,ab+2\,ac+2\,bc}} \geqq 0\)

Thiên tài

http://diendantoanhoc.net/topic/71619-cmr-fracab2a-bfraccb2c-bgeq-4/?setlanguage=1&langurlbits=topic/71619-cmr-fracab2a-bfraccb2c-bgeq-4/&langid=1

khó thế

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

\(\frac{a}{b}+\frac{c}{b}\le2\Leftrightarrow x+y\le2\)

VT=\(\frac{x+1}{2x-1}+\frac{y+1}{2y-1}\)

\(2VT=2+3\left(\frac{1}{2x-1}+\frac{1}{2y-1}\right)\ge2+\frac{3.4}{2\left(x+y\right)-2}\ge8\)

=> dpcm

a/ BĐT sai, cho \(a=b=c=2\) là thấy

b/ \(VT=\frac{a^4}{a^2+2ab}+\frac{b^4}{b^2+2bc}+\frac{c^4}{c^2+2ac}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a+b+c\right)^2}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)}{\left(a+b+c\right)^2}\)

\(VT\ge\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2}{3\left(a+b+c\right)^2}=\frac{1}{3}\left(a^2+b^2+c^2\right)\)

Dấu "=" xảy ra khi \(a=b=c\)

c/ Tiếp tục sai nữa, vế phải là \(\frac{3}{2}\) chứ ko phải \(2\), và hy vọng rằng a;b;c dương

\(VT=\frac{a^2}{abc.b+a}+\frac{b^2}{abc.c+b}+\frac{c^2}{abc.a+c}\ge\frac{\left(a+b+c\right)^2}{abc\left(a+b+c\right)+a+b+c}\)

\(VT\ge\frac{9}{3abc+3}\ge\frac{9}{\frac{3\left(a+b+c\right)^3}{27}+3}=\frac{9}{\frac{3.3^3}{27}+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta có:

\(a^3+b^3+b^3\ge3ab^2\) ; \(b^3+c^3+c^3\ge3bc^2\) ; \(c^3+a^3+a^3\ge3ca^2\)

Cộng vế với vế \(\Rightarrow a^3+b^3+c^3\ge ab^2+bc^2+ca^2\)

\(\frac{a^5}{b^2}+\frac{b^5}{c^2}+\frac{c^5}{a^2}=\frac{a^6}{ab^2}+\frac{b^6}{bc^2}+\frac{c^6}{ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{ab^2+bc^2+ca^2}\ge\frac{\left(a^3+b^3+c^3\right)^2}{a^3+b^3+c^3}=a^3+b^3+c^3\)

Ta có : \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow\frac{a+c}{ac}=\frac{2}{b}\Leftrightarrow b=\frac{2ac}{a+c}\)

\(\frac{a+b}{2a-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}=\frac{\frac{a^2+3ac}{a+c}}{\frac{2a^2}{a+c}}=\frac{a^2+3ac}{2a^2}=\frac{a+3c}{2a}\left(1\right)\)

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

Từ ( 1 ) ; ( 2 ) có : \(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+3c}{2a}+\frac{c+3a}{2c}=\frac{ac+3c^2+ac+3a^2}{2ac}=\frac{3\left(c^2+a^2\right)+2ac}{2ac}\)

Áp dụng BĐT Cauchy cho a ; c dương , ta có :

\(c^2+a^2\ge2ac\Rightarrow\frac{3\left(c^2+a^2\right)+2ac}{2ac}\ge\frac{3.2ac+2ac}{2ac}=4\)

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

Mà \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\) \(\Rightarrow\frac{2}{a}=\frac{2}{b}\Rightarrow a=b=c\)

Vậy ...

Theo giả thiết: \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{2}{\sqrt{ac}}\Leftrightarrow b^2\le ac\Leftrightarrow\frac{ac}{b^2}\ge1\)

Ta có: \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b\left(a+c\right)=2ac\Leftrightarrow2ac-bc=ab\Leftrightarrow2a-b=\frac{ab}{c}\)\(\Rightarrow\frac{a+b}{2a-b}=\frac{a+b}{\frac{ab}{c}}=\frac{ac+bc}{ab}=\frac{c}{b}+\frac{c}{a}\)(1)

Tương tự: \(\frac{b+c}{2c-b}=\frac{a}{c}+\frac{a}{b}\)(2)

Cộng từng vế hai đẳng thức (1), (2) và áp dụng Cô - si, ta được: \(\frac{a+b}{2a-b}+\frac{b+c}{2c-b}\ge\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+\frac{a}{b}\ge4\sqrt[4]{\frac{ca}{b^2}}\ge4\)

Đẳng thức xảy ra khi a = b = c

ta có \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Rightarrow b=\frac{2ac}{a+c}\)

thay b vào\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+3c}{2a}+\frac{c+3a}{2c}\)

                                                  \(=\frac{2ac+3\left(a^2+c^2\right)}{2ac}\ge\frac{2ac+6ac}{2ac}=4\)