\(1-\frac{a^2b}{2+a^2b}\ge1-\frac{a^2b}{3.\sqrt[3]{a^2b}}\)\(\rightarrow1-3\sqrt[3]{a^4b^2}=3.\sqrt[3]{ab.ab.a^2}\rightarrow.....\)

BĐT cần chứng minh tương đương với \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Áp dụng BĐT Cauchy ta có: \(2+a^2b=1+1+a^2b\ge3\sqrt[3]{a^2b}\)

Do đó ta được \(\frac{a^2b}{1+a^2b}\le\frac{a^2b}{3\sqrt[3]{a^2b}}=\frac{a\sqrt[3]{ab^2}}{3}\)

Hoàn toàn tương tự ta được \(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le\frac{a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}}{3}\)

Cũng theo BĐT Cauchy ta được \(\sqrt[3]{ab^2}\le\frac{a+b+b}{3}=\frac{a+2b}{3}\)

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

Tương tự cũng được \(a\sqrt[3]{ab^2}+b\sqrt[3]{bc^2}+c\sqrt[3]{ca}\le\frac{\left(a+b+c\right)^2}{3}=3\)

Từ đó ta được\(\frac{a^2b}{2+a^2b}+\frac{b^2c}{2+b^2c}+\frac{c^2a}{2+c^2a}\le1\)

Vậy BĐT được chứng minh. Dấu "=" xảy ra <=> a=b=c=1

1njfnjgjggnvfkgnbmvfvm 

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 )

Đặt \(a=x^2;b=y^2;c=z^2\)khi đó ta được xyz=1 và biểu thức P viết được thành

\(P=\frac{1}{x^2+2y^2+3}+\frac{1}{y^2+2x^2+3}+\frac{1}{z^2+2x^2+3}\)

Ta có \(x^2+y^2\ge2xy;y^2+1\ge2y\Rightarrow x^2+2y^2+3\ge2\left(xy+y+1\right)\)

Do đó ta được \(\frac{1}{x^2+2y^2+3}\le\frac{1}{2}\cdot\frac{1}{xy+y+1}\)

Chứng minh tương tự ta có:

\(\frac{1}{y^2+2z^2+3}\le\frac{1}{2}\cdot\frac{1}{yz+z+1};\frac{1}{z^2+2x^2+3}\le\frac{1}{2}\cdot\frac{1}{zx+z+1}\)

Cộng các vế BĐT trên ta được

\(P\le\frac{1}{2}\left(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}\right)\)

Ta cần chứng minh \(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+a+1}=1\)

Do xyz=1 nên ta được

\(\frac{1}{xy+y+1}+\frac{1}{yz+z+1}+\frac{1}{zx+x+1}=\frac{zx}{z+1+zx}+\frac{x}{1+zx+z}+\frac{1}{zx+x+1}=1\)

Từ đó ta được

\(P\le\frac{1}{2}\). Dấu "=" xảy ra <=> 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

Áp dụng bổ đề quen thuộc \(x^3+y^3\ge xy\left(x+y\right)\), ta được: \(\frac{1}{2a^3+b^3+c^3+2}=\frac{1}{\left(a^3+b^3\right)+\left(a^3+c^3\right)+2}\le\frac{1}{ab\left(a+b\right)+ac\left(a+c\right)+2}\)\(=\frac{bc}{ab^2c\left(a+b\right)+abc^2\left(a+c\right)+2bc}=\frac{bc}{b\left(a+b\right)+c\left(a+c\right)+2bc}\)\(\le\frac{bc}{ab+ac+4bc}=\frac{bc}{b\left(a+c\right)+c\left(a+b\right)+2bc}\)\(\le\frac{1}{9}\left(\frac{bc}{b\left(a+c\right)}+\frac{bc}{c\left(a+b\right)}+\frac{bc}{2bc}\right)=\frac{1}{9}\left(\frac{c}{a+c}+\frac{b}{a+b}+\frac{1}{2}\right)\)(1)

Tương tự, ta có: \(\frac{1}{a^3+2b^3+c^3+2}\le\frac{1}{9}\left(\frac{c}{b+c}+\frac{a}{a+b}+\frac{1}{2}\right)\)(2); \(\frac{1}{a^3+b^3+2c^3+2}\le\frac{1}{9}\left(\frac{b}{b+c}+\frac{a}{a+c}+\frac{1}{2}\right)\)(3)

Cộng theo vế ba bất đẳng thức (1), (2), (3), ta được: \(P\le\frac{1}{9}\left(1+1+1+\frac{3}{2}\right)=\frac{1}{2}\)

Vậy giá trị lớn nhất của P là \(\frac{1}{2}\)đạt được khi x = y = z = 1