hi kết bạn nha

3.Áp dụng BĐT \(\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)ta có

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

TT \(\frac{bc}{b+3c+2a}\le\frac{bc}{9}.\left(\frac{1}{b+a}+\frac{1}{2c}+\frac{1}{c+a}\right)\)

\(\frac{ca}{c+3a+2b}\le\frac{ac}{9}.\left(\frac{1}{a+b}+\frac{1}{2a}+\frac{1}{b+c}\right)\)

=> \(VT\le\frac{1}{18}\left(a+b+c\right)+\Sigma.\frac{1}{9}.\left(\frac{bc}{a+c}+\frac{ba}{a+c}\right)=\frac{1}{18}\left(a+b+c\right)+\frac{1}{9}\left(a+b+c\right)=\frac{1}{6}\left(a+b+c\right)\)

Dấu bằng xảy ra khi a=b=c

2. Chuẩn hóa \(a+b+c=3\)

=> \(ab+bc+ac\le3\)

=> \(c^2+3\ge\left(a+c\right)\left(b+c\right)\)

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

=> \(VT\le\Sigma\frac{1}{2}\left(\frac{ab}{a+c}+\frac{bc}{a+c}\right)=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=1

1. Ta có \(\sqrt{b^3+1}=\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\frac{1}{2}\left(b^2+2\right)\)

=> \(\frac{a}{\sqrt{b^3+1}}\ge\frac{2a}{2+b^2}=\frac{2a+ab^2-ab^2}{2+b^2}=a-\frac{2ab^2}{b^2+b^2+4}\)

Lại có \(b^2+b^2+4\ge3\sqrt[3]{b^4.4}\)

=> \(\frac{a}{\sqrt{b^3+1}}\ge a-\frac{2ab^2}{3\sqrt[3]{b^4.4}}=a-\frac{2}{3}.a.\sqrt[3]{\frac{b^2}{4}}\)

Mà \(\sqrt[3]{\frac{b^2}{4}.1}=\sqrt[3]{\frac{b}{2}.\frac{b}{2}.1}\le\frac{1}{3}\left(b+1\right)\)

=>\(\frac{a}{\sqrt[3]{b^3+1}}\ge a-\frac{2}{3}.a.\frac{1}{3}\left(b+1\right)=\frac{7a}{9}-\frac{2}{9}ab\)

Khi đó

\(VT\ge\frac{7}{9}\left(a+b+c\right)-\frac{2}{9}\left(ab+bc+ac\right)\)

Mà \(ab+bc+ac\le\frac{1}{3}\left(a+b+c\right)^2=12\)

=> \(VT\ge\frac{7}{9}.6-\frac{2}{9}.12=2\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c=2

Trả lời hộ mình đi

Áp dụng bđt Cauchy, ta có:

\(\sqrt{\frac{a}{bc}}\)+\(\sqrt{\frac{b}{ca}}\)≥ \(2\sqrt{\sqrt{\frac{ab}{abc^2}}}\)= \(2\sqrt{\sqrt{\frac{1}{c^2}}}\)= \(2\sqrt{\frac{1}{c}}\) (vì c>0)

Tương tự: \(\sqrt{\frac{b}{ca}}\)+\(\sqrt{\frac{c}{ab}}\)≥ \(2\sqrt{\frac{1}{a}}\)

                \(\sqrt{\frac{c}{ab}}\)+\(\sqrt{\frac{a}{bc}}\)≥ \(2\sqrt{\frac{1}{b}}\)

Cộng vế theo vế của các bđt với nhau, ta có: \(2\)\(\left(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\right)\text{≥}\)\(2\left(\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\right)\)

                                                             <=> \(\sqrt{\frac{a}{bc}}+\sqrt{\frac{b}{ca}}+\sqrt{\frac{c}{ab}}\text{≥}\)\(\sqrt{\frac{1}{a}}+\sqrt{\frac{1}{b}}+\sqrt{\frac{1}{c}}\)(đpcm)

Dấu "=" xảy ra <=> a = b = c

bài 1

<=> \(\frac{bc}{a\left(a+b+c\right)+bc}\)

sử dụng tiếp cauchy sharws

Bài 2: đặt a=x/y, b=y/x, c=z/x

a) Ta có BĐT:

\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\ge\left(a+b\right)ab\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b+c\right)}\)

Tương tự cho 2 bất đẳng thức còn lại rồi cộng theo vế:

\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}\)

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

Khi \(a=b=c\)

cảm ơn ạ

câu 1 . Theo bđt côsi ta có \(a^3+b^3\ge ab(a+b)\)

\(\Rightarrow\frac{1}{a^3+b^3+abc}\le\frac{1}{ab(a+b)+abc}=\frac{1}{ab(a+b+c)}=\frac{c}{abc(a+b+c)}\)

tương tự \(\frac{1}{b^3+c^3+abc}\le\frac{a}{abc(a+b+c)}\)và\(\frac{1}{a^3+c^3+abc}\le\frac{b}{abc(a+b+c)}\)

Cộng vế theo vế ta có  \(\frac{1}{b^3+c^3+abc}+\frac{1}{b^3+a^3+abc}+\frac{1}{a^3+c^3+abc}\le\frac{a+b+c}{abc(a+b+c)}=\frac{1}{abc}\)

\(\RightarrowĐPCM\)

Bài 1:

Đặt \(a^2=x;b^2=y;c^2=z\)

Ta có:\(\sqrt{\frac{x}{x+y}}+\sqrt{\frac{y}{y+z}}+\sqrt{\frac{z}{z+x}}\le\frac{3}{\sqrt{2}}\)

Áp dụng BĐT cô si ta có:

\(\sqrt{\frac{x}{x+y}}=\frac{1}{\sqrt{2}}\sqrt{\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}\frac{3\left(x+z\right)}{2\left(x+y+z\right)}}\)

\(\le\frac{1}{2\sqrt{2}}\left[\frac{4x\left(x+y+z\right)}{3\left(x+y\right)\left(x+z\right)}+\frac{3\left(x+z\right)}{2\left(x+y+z\right)}\right]\)

Tương tự với \(\sqrt{\frac{y}{y+z}}\)và \(\sqrt{\frac{z}{z+x}}\)

Cộng lại ta được:

\(\frac{\sqrt{2}}{3}\left[\frac{x\left(x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{y\left(x+y+z\right)}{\left(y+z\right)\left(y+x\right)}+\frac{z\left(x+y+z\right)}{\left(z+x\right)\left(z+y\right)}\right]+\frac{3}{2\sqrt{2}}\le\frac{3}{2\sqrt{2}}\)

Sau đó bình phương hai vế rồi

\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)đẳng thức đúng

Vậy...

Bài 2:

Trước hết ta chứng minh bất đẳng thức sau:

\(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\le\frac{1}{3}\)

Nhân cả hai vế bđt với 4(a+b+c)4(a+b+c) rồi thu gọn ta được bđt sau: 

\(\frac{4a\left(a+b+c\right)}{4a+4b+c}+\frac{4b\left(a+b+c\right)}{4b+4c+a}+\frac{4c\left(a+b+c\right)}{4c+4a+b}\)\(\le\frac{4}{3}\left(a+b+c\right)\)

\(\left[\frac{4a\left(a+b+c\right)}{4a+4b+}-a\right]+\left[\frac{4b\left(a+b+c\right)}{4b+4c+a}-b\right]+\left[\frac{4c\left(a+b+c\right)}{4c+4a+b}-c\right]\le\frac{a+b+c}{3}\)

\(\frac{ca}{4a+4b+c}+\frac{ab}{4b+4c+a}+\frac{bc}{4c+4a+b}\le\frac{a+b+c}{9}\)

Áp dụng bđt cauchy-Schwarz ta có \(\frac{ca}{4a+4b+c}=\frac{ca}{\left(2b+c\right)+2\left(2a+b\right)}\)\(\le\frac{ca}{9}\left(\frac{1}{2b+c}+\frac{2}{2a+b}\right)\)

Từ đó ta có:

\(\text{∑}\frac{ca}{4a+4b+c}\le\frac{1}{9}\text{∑}\left(\frac{ca}{2b+c}+\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ca}{2a+b}\right)\)\(=\frac{1}{9}\left(\text{ ∑}\frac{ca}{2b+c}+\text{ ∑}\frac{2ab}{2b+c}\right)=\frac{a+b+c}{9}\)

Đặt VT=A rồi áp dụng bđt cauchy-Schwarz cho VT ta có 

\(T^2\le3\left(\frac{a}{4a+4b+c}+\frac{b}{4b+4c+a}+\frac{c}{4c+4a+b}\right)\)\(\le3\cdot\frac{1}{3}=1\Leftrightarrow T\le1\)

Dấu = xảy ra khi a=b=c 

c bạn tự làm nhé mình mệt rồi :D

- Ôi má ơi, má patient dử dậy :)