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

\(\dfrac{a}{2b+a}+\dfrac{b}{2c+b}+\dfrac{c}{2a+b}=\dfrac{a^2}{2ab+a^2}+\dfrac{b^2}{2bc+b^2}+\dfrac{c^2}{2ac+bc}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ac}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2}=1\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Vậy ...

Ta có: \(\dfrac{a^3}{a^2+2b^2}=a-\dfrac{2ab^2}{a^2+2b^2}\ge a-\dfrac{2ab^2}{3\sqrt[3]{a^2b^4}}=a-\dfrac{2}{3}\sqrt[3]{ab^2}\ge a-\dfrac{2}{9}\left(a+b+b\right)=a-\dfrac{2}{9}\left(a+2b\right)\) Chứng minh tương tự ta được:

\(\dfrac{b^3}{b^2+2c^2}\ge b-\dfrac{2}{9}\left(b+2c\right);\dfrac{c^3}{c^2+2a^2}\ge c-\dfrac{2}{9}\left(c+2a\right)\)

\(\Rightarrow\dfrac{a^3}{a^2+2b^2}+\dfrac{b^3}{b^2+2c^2}+\dfrac{c^3}{c^2+2a^2}\ge a+b+c-\dfrac{2}{9}\left(a+2b+b+2c+c+2a\right)=a+b+c-\dfrac{2}{9}\left(3a+3b+3c\right)=\dfrac{1}{3}\left(a+b+c\right)\ge\dfrac{1}{3}\cdot3\sqrt[3]{abc}=1\)Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)

cho a, b, c là các số thực dương thỏa mạn abc=1 chứng minh rằng a/(2b+a) +b/(2c+b)+c/(2a+c)>=1

Trước hết ta chứng minh các bđt : \(a^7+b^7\ge a^2b^2\left(a^3+b^3\right)\left(1\right)\)

Thật vậy:

\(\left(1\right)\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\left(a^4+a^3b+a^2b^2+ab^3+b^4\right)\ge0\)(luôn đúng)

Lại có : \(a^3+b^3+1\ge ab\left(a+b+1\right)\)

\(\Leftrightarrow a^3+b^3+abc\ge ab\left(a+b+1\right)\)

mà \(a^3+b^3\ge ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+1\right)\)(luôn đúng)

Áp dụng các bđt trên vào bài toán ta có

 ∑\(\frac{a^2b^2}{a^7+a^2b^2+b^7}\le\)∑\(\frac{a^2b^2}{a^3b^3\left(a+b+c\right)}\le\)∑\(\frac{a+b+c}{a+b+c}=1\)

Bất đẳng thức được chứng minh

Dấu "=" xảy ra khi a=b=c=1

Em xem lại dòng thứ 4 và giải thích lại giúp cô với! ko đúng hoặc bị nhầm

chứng minh bđt "Lại có" ạ

Ta có: \(2a+b^2=2a\left(a+b+c\right)+b^2=b^2+2a^2+2ab+2ac\)

\(\ge4ab+2ac+a^2\)

\(\Rightarrow\frac{a}{2a+b^2}\le\frac{a}{4ab+2ac+a^2}=\frac{1}{4b+2c+a}\)

\(\le\frac{1}{49}.\frac{49}{4b+2c+a}=\frac{1}{49}.\frac{\left(4+2+1\right)^2}{4b+2c+a}\)

\(\le\frac{1}{49}\left(\frac{16}{4b}+\frac{4}{2c}+\frac{1}{a}\right)=\frac{1}{49}\left(\frac{4}{b}+\frac{2}{c}+\frac{1}{a}\right)\)

CMTT: \(\frac{b}{2b+c^2}\le\frac{1}{49}\left(\frac{4}{c}+\frac{2}{a}+\frac{1}{b}\right);\frac{c}{2c+a^2}\le\frac{1}{49}\left(\frac{4}{a}+\frac{2}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{2a+b^2}+\frac{b}{2b+c^2}+\frac{c}{2c+a^2}\le\frac{1}{7}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)( đpcm )

Bài này chả khó với lại đầy người đăng rồi

Ta có: \(a^2+b^2\ge2ab\) và \(b^2+1\ge2b\)

\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2=2\left(ab+b+1\right)\)

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

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

Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\) ta có:

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

\(=\frac{1}{2}\left(\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\right)\)

\(=\frac{1}{2}\left(\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}\right)\left(abc=1\right)\)

\(=\frac{1}{2}\left(\frac{ac+a+1}{ac+a+1}\right)=\frac{1}{2}=VP\) (ĐPCM)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

Đẳng thức xảy ra khi a = b = c = 1/3

Bài này không khó! Sao lại được vào câu hỏi hay?

tk cho mình nhìu nhìu nha! Gõ máy mệt quá!

Ta có: \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)\(;b^2+1\ge2\sqrt{b^2\cdot1}=2b\)

\(\Rightarrow a^2+2b^2+3\ge2ab+2b+2=2\left(ab+b+1\right)\)

\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2}\left(ab+b+1\right)\left(1\right)\). Tương tự ta có:

\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\left(bc+c+1\right)\left(2\right);\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\left(ac+a+1\right)\left(3\right)\)

Cộng theo vế của (1);(2) và (3) ta có:

\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\)

\(\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab+b+1}+\frac{b}{ab+b+1}\right)=\frac{1}{2}\) (vì abc=1)

Suy ra Đpcm. Dấu "=" khi a=b=c=1

Ta có: 

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

Tương tự CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1}\) và \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)

\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)

\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)=\frac{1}{2}\cdot1=\frac{1}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

A=\(\frac{1}{a^2+2b^2+3}\)+\(\frac{1}{b^2+2c^2+3}\)+\(\frac{1}{c^2+2a^2+3}\)

ta có: \(\frac{1}{a^2+2b^2+3}\)=\(\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\)\(\le\)\(\frac{1}{2\left(ab+b+1\right)}\)

vì : a²+b²\(\ge\)2\(\sqrt{a^2b^2}\)=2ab

b²+1\(\ge\)2\(\sqrt{b^2x1}\)=2b

cmtt => A\(\le\)\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{1}{bc+c+1}\)+\(\frac{1}{ca+a+1}\))

=\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab^2c+abc+ab}\)+\(\frac{b}{cba+ab+b}\))

=\(\frac{1}{2}\)x (\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab+b+1}\)+\(\frac{b}{ab+b+1}\))=\(\frac{1}{2}\)x\(\frac{ab+b+1}{ab+b+1}\)=\(\frac{1}{2}\)

dấu "=" xảy ra <=> a=b=c=1

+ 1a2+2b2+3=1(a2+b2)+(b2+1)+2≤12(ab+b+1)1a2+2b2+3=1(a2+b2)+(b2+1)+2≤12(ab+b+1) . Dấu "=" ⇔a=b=1⇔a=b=1

+ Tương tự : 1b2+2c2+3≤12(bc+c+1)1b2+2c2+3≤12(bc+c+1). Dấu "=" ⇔b=c=1⇔b=c=1

1c2+2a2+3≤12(ca+a+1)1c2+2a2+3≤12(ca+a+1). Dấu "=" c=a=1c=a=1

Do đó : VT≤12(1ab+b+1+1bc+c+1+1ca+a+1)=12(1ab+b+1+ababc⋅b+abc+ab+babc+ab+b)VT≤12(1ab+b+1+1bc+c+1+1ca+a+1)=12(1ab+b+1+ababc⋅b+abc+ab+babc+ab+b)

=12(1ab+b+1+abab+b+1+bab+b+1)=12=12(1ab+b+1+abab+b+1+bab+b+1)=12

Dấu "=" ⇔a=b=c=1