BĐT cần  chứng minh tương đương với :

\(\left(a^2b+b^2c+c^2a\right)\left(2+\frac{1}{a^2b^2c^2}\right)\ge9\)

\(\Leftrightarrow2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge9\)

Áp dụng BĐT Cô-si cho 3 số dương ,ta có :

\(a^2b+a^2b+\frac{1}{ab^2}\ge3\sqrt[3]{a^2b.a^2b.\frac{1}{ab^2}}=3a\)

tương tự :  \(b^2c+bc^2+\frac{1}{bc^2}\ge3b\), \(\left(c^2a+ca^2+\frac{1}{ca^2}\right)\ge3c\)

Cộng 3 BĐT trên theo vế, ta được :

\(2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge3\left(a+b+c\right)=9\)

Dấu "=" xảy ra khi a = b = c = 1

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

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

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

Tương tự ...

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

\(=\dfrac{1}{2}\left(\dfrac{c}{abc+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{ca.bc+a.bc+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c}{1+bc+c}+\dfrac{1}{bc+c+1}+\dfrac{bc}{c+1+bc}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{c+1+bc}{1+bc+c}\right)=\dfrac{1}{2}\)

\(P_{max}=\dfrac{1}{2}\) khi \(a=b=c=1\)

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 \(\Leftrightarrow a=b=c=1\)

mong các thầy cô giúp em giải bài này với ạ