Mình chỉ làm sơ sơ, có gì bạn sửa lại

Ta có: \(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\)

Đặt  a  ;   b và c = 2 . 

Thế số vào biểu thức ta có: 

\(\frac{2}{\sqrt{2^3+1}}+\frac{2}{\sqrt{2^3+1}}+\frac{2}{\sqrt{2^3+1}}\)

\(\Leftrightarrow\frac{2}{\left(2^3+1\right)^2}+\frac{2}{\left(2^3+1\right)^2}+\frac{2}{\left(2^3+1\right)^2}\)

\(\Leftrightarrow\frac{2}{\left(2^3+1\right)^2}.3\Leftrightarrow\frac{2}{\left(8+1\right)^2}.3\Leftrightarrow\frac{2}{9^2}\ge2\)

Ta có ĐPCM

Conan: bác mori ơi cháu biết hung thủ là ai rồi

Mouri : cái j , trẻ con đi chỗ khác chơi

Conan : hừ , lại phải dùng thuốc gây mê rồi ,  pặc

Mouri : á á :) , lại thế nữa rồi , á á 

Conan : thanh tra megure ơi bác mouri nói đã tìm ra hung thủ rồi

megure : Thật không Mori , anh đã tìm ra hung thủ rồi à 

Mouri : chính xác hung thủ chính là hắn :) 

dự đoán của Mouri a=b=c=2

áp dụng BDT cô si ta có

\(VT\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{\sqrt{b^3+1}+\sqrt{c^3+1}+\sqrt{a^3+1}}.\)

áp dụng BDT cô si dạng shinra " mẫu số" ta có   với Q= mẫu số

\(\sqrt{\left(b^3+1\right).9}\le\frac{b^3+1+9}{2}\)

\(\sqrt{\left(c^3+1\right).9}\le\frac{c^3+1+9}{2}\)

\(\sqrt{a^3+1.9}\le\frac{a^3+1+9}{2}\)

\(3Q\le\frac{1}{2}\left(a^3+b^3+c^3\right)+15.\)

có

\(a^3+8+8\ge3\sqrt[3]{a^32^32^3}=12a\)

\(b^3+8+8\ge12b\)

\(c^3+8+8\ge12c\)

\(a^3+b^3+c^3\ge72-48=24\)

\(3Q\le\frac{24}{2}+15=27\Leftrightarrow Q=9\)

thay vào VT ta được

\(VT\ge\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{9}\)

\(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\left(a+b+c\right)+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

\(VT\ge\frac{6+2\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}{9}\)

\(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\ge3\sqrt[3]{\sqrt{a^2b^2c^2}}=3\sqrt[3]{abc}\)

\(a+b+c\ge3\sqrt[3]{abc}\)

suy ra đươc  \(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}=a+b+c=6\)

\(VT\ge\frac{6+2\left(6\right)}{9}=2\)

dấu = xảy ra khi a=b=c=2

p/s đúng nhé

Áp dụng BĐT AM-GM và Cauchy-Schwarz ta có:

\(VT=Σ_{cyc}\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\geΣ_{cyc}\frac{a}{\sqrt{\frac{\left(b+1+b^2-b+1\right)^2}{4}}}\)

\(=Σ_{cyc}\frac{2a}{b^2+2}\)\(=Σ_{cyc}\frac{2a^2}{ab^2+2a}\ge\frac{2\left(a+b+c\right)^2}{Σ_{cyc}ab^2+2\left(a+b+c\right)}\)

Cần c.minh \(\frac{2\left(a+b+c\right)^2}{Σ_{cyc}ab^2+2\left(a+b+c\right)}\ge2\)\(\Leftrightarrow\frac{36}{Σ_{cyc}ab^2+12}\ge1\)

Hay \(ab^2+bc^2+ca^2\le24\)\(\Leftrightarrow\)\(\left(a+b+c\right)^3\ge9\left(ab^2+bc^2+ca^2\right)\left(☺\right)\)

\(VT_{\left(☺\right)}\ge3\left(a+b+c\right)\left(ab+bc+ac\right)\ge9\left(ab^2+bc^2+ca^2\right)\) (vì \(\left(Σa\right)^2\ge3\left(Σab\right)\))

\(\Leftrightarrow\left(a+b+c\right)\left(ab+ac+bc\right)\ge3\left(ab^2+bc^2+ca^2\right)\)

Tự c.m nốt gợi ý: \(a^2b+b^2c+c^2a-\)\(\left(ab^2+bc^2+ca^2\right)\)\(=\frac{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}{3}\)

Và \(3abc-\left(ab^2+bc^2+ca^2\right)=ab\left(c-b\right)+bc\left(a-c\right)+ac\left(b-a\right)\)

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

Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(VT=\Sigma\frac{a}{\sqrt{b^3+1}}=\Sigma\frac{a}{\sqrt{\left(b+1\right)\left(b^2-b+1\right)}}\)

\(\ge\Sigma\frac{a}{\frac{\left(b+1\right)+\left(b^2-b+1\right)}{2}}=\Sigma\frac{2a}{b^2+2}=\Sigma\left(a-\frac{ab^2}{b^2+2}\right)\)

\(=\Sigma\left(a-\frac{2ab^2}{b^2+b^2+4}\right)\ge\Sigma\left(a-\frac{2ab^2}{3\sqrt[3]{4b^4}}\right)\)\(=\Sigma\left[a-\frac{a\sqrt[3]{2b^2}}{3}\right]=\Sigma\left[a-\frac{a\sqrt[3]{2.b.b}}{3}\right]\)

\(\ge\Sigma\left[a-\frac{a\left(2+b+b\right)}{9}\right]\)\(=\left(a+b+c\right)-\frac{2\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)

\(=\frac{7\left(a+b+c\right)}{9}-\frac{2\left(ab+bc+ca\right)}{9}\)\(\ge\frac{7\left(a+b+c\right)}{9}-\frac{2.\frac{\left(a+b+c\right)^2}{3}}{9}=2\)

Đẳng thức xảy ra khi a = b = c = 2

ap dung bat dang thuc amgm

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

\(\Rightarrow\frac{a}{\sqrt{b^3+1}}\ge2.\frac{a}{b^2+2}\)

P=\(\frac{a}{\sqrt{b^3+1}}+\frac{b}{\sqrt{c^3+1}}+\frac{c}{\sqrt{a^3+1}}\ge2\left(\frac{a}{b^2+2}+\frac{b}{c^2+2}+\frac{c}{a^2+2}\right)\) \(\)  

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

tiep tuc ap dung bdt cauchy-swart dang phan thuc 

\(\ge2\frac{\left(a+b+c\right)^2}{a\left(b^2+2\right)+b\left(c^2+2\right)+c\left(a^2+2\right)}\)=