Áp dụng BĐT sau: \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) 

\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{b+c}\right)\). Lại có \(\frac{1}{b+c}\le\frac{1}{4b}+\frac{1}{4c}\)

\(\Rightarrow\frac{1}{2a+b+c}\le\frac{1}{4}\left(\frac{1}{2a}+\frac{1}{4b}+\frac{1}{4c}\right)\)

Tương tự: \(\frac{1}{a+2b+c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{2b}+\frac{1}{4c}\right);\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{4a}+\frac{1}{4b}+\frac{1}{2c}\right)\)

Cộng 3 BĐT trên theo vế, ta được:

\(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Thay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\)\(\Rightarrow\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\le1\)(đpcm).

Dấu "=" xảy ra <=> \(a=b=c=\frac{3}{4}.\)

Đặt \(\left(\frac{1}{a},\frac{1}{b},\frac{1}{c}\right)=\left(x,y,z\right)\)

\(x+y+z\ge\frac{x^2+2xy}{2x+y}+\frac{y^2+2yz}{2y+z}+\frac{z^2+2zx}{2z+x}\)

\(\Leftrightarrow x+y+z\ge\frac{3xy}{2x+y}+\frac{3yz}{2y+z}+\frac{3zx}{2z+x}\)

\(\frac{3xy}{2x+y}\le\frac{3}{9}xy\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}\right)=\frac{1}{3}\left(x+2y\right)\)

\(\Rightarrow\Sigma_{cyc}\frac{3xy}{2x+y}\le\frac{1}{3}\left[\left(x+2y\right)+\left(y+2z\right)+\left(z+2x\right)\right]=x+y+z\)

Dấu "=" xảy ra khi x=y=z

Từ \(ab+bc+ca=5abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=5\)

Áp dụng BĐT Bu-nhi-a-cốp-xki ta có :

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

\(\Rightarrow\frac{2}{a}+\frac{2}{b}+\frac{1}{c}\ge\frac{25}{2a+2b+c}\)

Tương tự ta có :

\(\frac{2}{b}+\frac{2}{c}+\frac{1}{a}\ge\frac{25}{2b+2c+a}\)

\(\frac{2}{a}+\frac{1}{b}+\frac{2}{c}\ge\frac{25}{2a+b+2c}\)

Cộng từng vế BĐT ta thu được :

\(\frac{5}{a}+\frac{5}{b}+\frac{5}{c}\ge25P\)

\(\Leftrightarrow P\le\frac{5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}{25}=1\)

Vậy BĐT đã được chứng minh . Dấu \("="\) xảy ra khi \(a=b=c=\frac{3}{5}\)

Cho $a, b, c > 0$. Đặt $P = \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c}$.

Ta cần chứng minh: $\frac{3}{5} \le P < 1$.

* Chứng minh $P \ge \frac{3}{5}$:

Áp dụng bất đẳng thức Cauchy-Schwarz dạng phân thức:

$P = \frac{a^2}{a(a+2b+2c)} + \frac{b^2}{b(2a+b+2c)} + \frac{c^2}{c(2a+2b+c)}$

$\ge \frac{(a + b + c)^2}{a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)}$

Rút gọn mẫu thức:

$a(a+2b+2c) + b(2a+b+2c) + c(2a+2b+c)$

$= a^2 + 2ab + 2ac + 2ab + b^2 + 2bc + 2ac + 2bc + c^2$

$= a^2 + b^2 + c^2 + 4ab + 4bc + 4ca$

$= (a + b + c)^2 + 2(ab + bc + ca)$

Ta có bất đẳng thức quen thuộc: $ab + bc + ca \le \frac{(a + b + c)^2}{3}$

Do đó mẫu thức:

$(a + b + c)^2 + 2(ab + bc + ca) \le (a + b + c)^2 + 2 \cdot \frac{(a + b + c)^2}{3} = \frac{5}{3}(a + b + c)^2$

Suy ra: $P \ge \frac{(a + b + c)^2}{\frac{5}{3}(a + b + c)^2} = \frac{3}{5}$

Dấu "=" xảy ra khi $a = b = c$.

Vì $a, b, c > 0$ nên:

$a + 2b + 2c > a + b + c \Rightarrow \frac{a}{a+2b+2c} < \frac{a}{a+b+c}$

$2a + b + 2c > a + b + c \Rightarrow \frac{b}{2a+b+2c} < \frac{b}{a+b+c}$

$2a + 2b + c > a + b + c \Rightarrow \frac{c}{2a+2b+c} < \frac{c}{a+b+c}$

Cộng từng vế của ba bất đẳng thức trên:

$P < \frac{a}{a+b+c} + \frac{b}{a+b+c} + \frac{c}{a+b+c}$

$\Rightarrow P < \frac{a + b + c}{a + b + c} = 1$

Vậy $\frac{3}{5} \le \frac{a}{a+2b+2c} + \frac{b}{2a+b+2c} + \frac{c}{2a+2b+c} < 1$ (điều phải chứng minh).

3.

\(5a^2+2ab+2b^2=\left(a^2-2ab+b^2\right)+\left(4a^2+4ab+b^2\right)\)

\(=\left(a-b\right)^2+\left(2a+b\right)^2\ge\left(2a+b\right)^2\)

\(\Rightarrow\sqrt{5a^2+2ab+2b^2}\ge2a+b\)

\(\Rightarrow\frac{1}{\sqrt{5a^2+2ab+2b^2}}\le\frac{1}{2a+b}\)

Tương tự \(\frac{1}{\sqrt{5b^2+2bc+2c^2}}\le\frac{1}{2b+c};\frac{1}{\sqrt{5c^2+2ca+2a^2}}\le\frac{1}{2c+a}\)

\(\Rightarrow P\le\frac{1}{2a+b}+\frac{1}{2b+c}+\frac{1}{2c+a}\)

\(\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\right)\)

\(=\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{3}.\sqrt{3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{\sqrt{3}}{3}\)

\(\Rightarrow MaxP=\frac{\sqrt{3}}{3}\Leftrightarrow a=b=c=\sqrt{3}\)

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

Tương tự: \(\frac{3}{b+2c}\le\frac{1}{3}\left(\frac{1}{b}+\frac{2}{c}\right)\) ; \(\frac{3}{c+2a}\le\frac{1}{3}\left(\frac{1}{c}+\frac{2}{a}\right)\)

Cộng vế với vế:

\(3\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\le\frac{1}{3}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

moi nguoi oi hom truoc minh hoc tap hop cac so TN do thi co cua minh day nhu sau 

vd: A={xeN/3<x<9}

thi minh liet ke ra la A=4,5,6,7,8 nhung sua bai lai ko dung 

co sua nhu vay A=3,4,5,6,7,8

ko biet hay sai mong ae giup minh

Áp dụng BĐT Cô-si \(ab\le\frac{\left(a+b\right)}{4}^2\)

=> \(\left(2a+b\right)\left(2c+b\right)\le\frac{4\left(a+b+c\right)^2}{4}=\left(a+b+c\right)^2\)

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

Mấy cái kia làm tương tự cậu nhé 

Dấu "=" xảy ra khi và chỉ khi a=b=c=1