n² + 3 ⋮ n + 1 <=> ( n² + 1 ) + 2 ⋮ n + 1

=> ( n + 1 ).( n - 1 ) + 2 ⋮ n + 1

Vì ( n + 1 ).( n - 1 ) ⋮ n + 1 , để ( n + 1 )( n - 1 ) + 2 ⋮ n + 1 thì 2 ⋮ n + 1 => n + 1 ∈ Ư ( 2 ) = { + 1 ; + 2 }

Ta có : n + 1 = 1 => n = 0 ( thỏa mãn )

            n + 1 = - 1 => n = - 2 ( TM )

            n + 1 = 2 => n = 1 ( TM )

           n + 1 = - 2 => n = - 3 ( TM )

Vậy n = { - 3 ; - 2 ; 0 ; 1 }

{-5;-3;-2;0;1;3} , ủng hộ mk nha

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\ge\dfrac{9}{ab+bc+ac}\)

\(\Rightarrow VT\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ac}\)

\(\Leftrightarrow VT\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}+\dfrac{7}{ab+ac+bc}\)

Theo hệ quả của bất đẳng thức Cauchy

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

\(\Rightarrow\dfrac{7}{ab+bc+ac}\ge21\) (1)

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ac}+\dfrac{1}{ab+bc+ac}\ge\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ac\right)}=9\) (2)

Từ (1) và (2)

\(\Rightarrow VT\ge21+9=30\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow VT\ge3\sqrt[3]{\dfrac{1}{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}=\dfrac{3}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}\)

Chứng minh rằng \(\dfrac{3}{\sqrt[3]{\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)}}\ge\dfrac{3}{1+xyz}\)

\(\Leftrightarrow\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)\le\left(1+xyz\right)^3\)

Áp dụng bất đẳng thức Holder

\(\Rightarrow\left(1+x^3\right)\left(1+y^3\right)\left(1+z^3\right)\ge\left(1+xyz\right)^3\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=1\)

Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)

Áp dụng bất đẳng thức Cauchy

Ta có

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

\(\sqrt{c^2+ab}\le\dfrac{1}{2}\left(b+c+\dfrac{c^2+ab}{b+c}\right)\)

\(\Rightarrow VT\le\dfrac{1}{2}\left(a+2b+3c+\dfrac{a^2+bc}{a+c}+\dfrac{ab+ac+b^2+c^2}{b+c}\right)\)

\(\Leftrightarrow VT\le\dfrac{1}{2}\left(2a+2b+3c+\dfrac{a^2+bc}{a+c}+\dfrac{b^2+c^2}{b+c}\right)\)

Ta sẽ chứng minh rằng \(\dfrac{1}{2}\left(2a+2b+3c+\dfrac{a^2+bc}{a+c}+\dfrac{b^2+c^2}{b+c}\right)\le\dfrac{3}{2}\left(a+b+c\right)\)

\(\Leftrightarrow\dfrac{a^2+bc}{a+c}+\dfrac{b^2+c^2}{b+c}\le a+b\)

Hiển nhiên rằng \(\dfrac{a^2+bc}{a+c}\le a\)

\(\Leftrightarrow a^2+bc\le a^2+ac\Leftrightarrow a\ge b\) ( đúng với giả sử )

Chứng minh tương tự ta có \(\dfrac{b^2+c^2}{b+c}\le b\Leftrightarrow b^2+c^2\le b^2+bc\Leftrightarrow c\le b\) ( đúng vời giả sử )

\(\Rightarrow\dfrac{a^2+bc}{a+c}+\dfrac{b^2+c^2}{b+c}\le a+b\left(đpcm\right)\)

\(\Rightarrowđpcm\)

\(VT=\dfrac{a}{b\left(b^2+a\right)}+\dfrac{b}{c\left(c^2+b\right)}+\dfrac{c}{a\left(a^2+c\right)}\)

\(VT=\dfrac{a+b^2-b^2}{b\left(b^2+a\right)}+\dfrac{b+c^2-c^2}{c\left(c^2+b\right)}+\dfrac{c+a^2-a^2}{a\left(a^2+c\right)}\)

\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)

\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{b}{b^2+a}+\dfrac{c}{c^2+b}+\dfrac{a}{a^2+c}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{b}{b^2+a}\le\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)

Thiết lập tương tự và thu lại tao có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}+\dfrac{1}{\sqrt{c}}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\sqrt{\dfrac{1}{a}}\le\dfrac{\dfrac{1}{a}+1}{2}\)

Tương tự ta có

\(\sqrt{\dfrac{1}{b}}\le\dfrac{\dfrac{1}{b}+1}{2};\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{c}+1}{2}\)

Thu lại ta có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\right)\)

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)

\(\Rightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)-\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}-\dfrac{3}{4}=\dfrac{3}{2}\)

\(\Rightarrow VT\ge\dfrac{3}{2}\left(đpcm\right)\)

Dấu " = " xảy ra khi \(a=b=c=1\)

\(VT=\dfrac{a}{b\left(b^2+a\right)}-\dfrac{b}{c\left(c^2+b\right)}-\dfrac{c}{a\left(a^2+c\right)}\)

\(VT=\dfrac{a+b^2-b^2}{b\left(b^2+a\right)}-\dfrac{b+c^2-c^2}{c\left(c^2+b\right)}-\dfrac{c+a^2-a^2}{a\left(a^2+c\right)}\)

\(VT=\dfrac{1}{b}-\dfrac{b}{b^2+a}+\dfrac{1}{c}-\dfrac{c}{c^2+b}+\dfrac{1}{a}-\dfrac{a}{a^2+c}\)

\(VT=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\left(\dfrac{b}{b^2+a}+\dfrac{c}{c^2+b}+\dfrac{a}{a^2+c}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{b}{b^2+a}\ge\dfrac{b}{2b\sqrt{a}}=\dfrac{1}{2\sqrt{a}}\)

Thiết lập tương tự và thu lại tao có

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\right)\)

Áp dụng bất đẳng thức Cauchy

\(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\le\dfrac{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3}{2}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{2}\left(\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}}+\sqrt{\dfrac{1}{c}}\right)\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)

\(\Rightarrow VT\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+3\right)\)

\(\Leftrightarrow VT\ge\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy dạng phân thức

\(\Rightarrow\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\ge\dfrac{3}{4}.\dfrac{9}{a+b+c}+\dfrac{3}{4}=3\)

\(\Rightarrow VT\ge3\left(đpcm\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow VT\ge3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\)

Chứng minh \(3\sqrt[6]{\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)}}\ge3\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\left(c+ab\right)\left(a+bc\right)\le\dfrac{\left(c+a+ab+bc\right)^2}{4}=\dfrac{\left[b\left(a+c\right)+c+a\right]^2}{4}=\dfrac{\left(b+1\right)^2\left(c+a\right)^2}{4}\)

Thiết lập tương tự và thu lại ta có

\(\Rightarrow\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\dfrac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(a+1\right)^2\left(c+1\right)^2}{64}\)

\(\Rightarrow64\left(c+ab\right)^2\left(a+bc\right)^2\left(b+ac\right)^2\le\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\left(b+1\right)^2\left(c+1\right)^2\left(a+1\right)^2\)

\(\Leftrightarrow8\left(c+ab\right)\left(a+bc\right)\left(b+ac\right)\le\left(a+b\right)\left(b+c\right)\left(c+a\right)\left(b+1\right)\left(c+1\right)\left(a+1\right)\)

Cần chứng minh rằng \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le8\)

Áp dụng bất đẳng thức Cauchy \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\le\left(\dfrac{3+3}{3}\right)^3=8\left(đpcm\right)\)

\(\Rightarrowđpcm\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\Sigma\dfrac{1}{2x+3y+3z}\le\Sigma\dfrac{1}{16}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)

\(\Rightarrow P\le\dfrac{4}{16}\Sigma\left(\dfrac{1}{x+y}\right)=\dfrac{2017}{4}\)

Dấu " = " xảy ra khi \(x=y=z=\dfrac{3}{4034}\)