Bài 1:

\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)

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

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

\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)

Tương tự cho 2 BĐT còn lại ta cũng có: 

\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)

Cộng theo vế 3 BĐT trên ta có: 

\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

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

Đẳng thức xảy ra khi \(a=b=c\)

Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2

Bài 2/

\(\frac{bc}{a^2b+a^2c}+\frac{ca}{b^2c+b^2a}+\frac{ab}{c^2a+c^2b}\)

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

\(=\frac{b^2c^2}{ab+ac}+\frac{c^2a^2}{bc+ba}+\frac{a^2b^2}{ca+cb}\)

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)

\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)

Dấu =  xảy ra khi \(a=b=c=1\)

Dat \(P=\frac{a}{\sqrt{2b^2+2c^2-a^2}}+\frac{b}{\sqrt{2c^2+2a^2-b^2}}+\frac{c}{\sqrt{2a^2+2b^2-c^2}}\)

Ta co:

\(\frac{a}{\sqrt{2b^2+2c^2-a^2}}=\frac{\sqrt{3}a^2}{\sqrt{3a^2\left(2b^2+2c^2-a^2\right)}}\ge\frac{\sqrt{3}a^2}{a^2+b^2+c^2}\)

Tuong tu:

\(\frac{b}{\sqrt{2c^2+2a^2-b^2}}\ge\frac{\sqrt{3}b^2}{a^2+b^2+c^2}\)

\(\frac{c}{\sqrt{2a^2+2b^2-c^2}}\ge\frac{\sqrt{3}c^2}{a^2+b^2+c^2}\)

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

Dau '=' xay ra khi \(a=b=c\)

Áp dụng BĐT Cauchy - Schwarz và BĐT phụ \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\)

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

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

\(\le\frac{3}{4}\left(\frac{a}{b+a}+\frac{a}{c+a}+\frac{b}{b+c}+\frac{b}{b+a}+\frac{c}{c+a}+\frac{c}{c+b}\right)\)

\(=\frac{3}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)=\frac{9}{4}\)

\(\Rightarrow M\le\frac{3}{2}\)

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

ko hỉu

ÁP DỤNG BĐT COSI TA CÓ :\(\sqrt{\frac{a}{b+c+2a}}\le\frac{a}{b+c+2a}+\frac{1}{4}\)

                                            \(\sqrt[]{\frac{b}{a+c+2b}}\le\frac{b}{a+c+2b}+\frac{1}{4}\)

                                            \(\sqrt[]{\frac{c}{a+b+2c}}\le\frac{c}{a+b+2c}+\frac{1}{4}\)

ĐẶT A=\(\sqrt[]{\frac{a}{b+c+2a}}+\sqrt[]{\frac{b}{a+c+2b}}+\sqrt[]{\frac{c}{a+b+2c}}\)

            \(\le\frac{a}{b+c+2a}+\frac{b}{a+c+2b}+\frac{c}{a+b+2c}+\frac{3}{4}\)

        ÁP DỤNG BĐT :\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

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

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

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

  \(\Rightarrow A\le\frac{1}{4}\left(\frac{a}{a+b}+\frac{a}{a+c}+\frac{b}{b+c}+\frac{b}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}\right)+\frac{3}{4}\)

 \(\Rightarrow A\le\frac{1}{4}\left(\frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}\right)+\frac{3}{4}\)

\(\Rightarrow A\le\frac{1}{4}\left(1+1+1\right)+\frac{3}{4}\)

\(\Rightarrow A\le\frac{3}{2}\)

DẤU = XẢY RA\(\Leftrightarrow a=b=c\)

Một lời giải khác: 

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

\(\le\left[\Sigma a\left(a+2c+b\right)\right]\left[\Sigma\frac{1}{\left(a+2c+b\right)\left(b+c+2a\right)}\right]=\Sigma\frac{a^2+3ab}{\left(a+2c+b\right)\left(b+c+2a\right)}\)

\(=\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\)

Cần chứng minh \(\frac{4\left(\Sigma a^2+3\Sigma ab\right)\left(\Sigma a\right)}{\Pi\left(a+2c+b\right)}\le\frac{9}{4}\)

Chịu khó quy đồng :V

\(A=\frac{a\sqrt{a}}{\sqrt{a+b+2c}}+\frac{b\sqrt{b}}{\sqrt{b+c+2a}}+\frac{c\sqrt{c}}{\sqrt{c+a+2b}}\)

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

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

Xét: \(2\left(\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\right)\)

\(=\sqrt{4a\left(a+b+2c\right)}+\sqrt{4b\left(b+c+2a\right)}+\sqrt{4c\left(c+a+2b\right)}\)

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

\(\Rightarrow\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}\le2\left(a+b+c\right)\)

\(\Rightarrow\frac{\left(a+b+c\right)^2}{\sqrt{a\left(a+b+2c\right)}+\sqrt{b\left(b+c+2a\right)}+\sqrt{c\left(c+a+2b\right)}}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c=1\)

Ta có \(\frac{\sqrt{ab^2c^3}}{b+c}\le\frac{\sqrt{ab^2c^3}}{2\sqrt{bc}}=\frac{1}{2}.\sqrt{ac.bc}\)

Mà \(\frac{1}{2}\sqrt{ac.cb}\le\frac{1}{4}\left(ac+cb\right)\)\(\Rightarrow\frac{\sqrt{ab^2c^3}}{b+c}\le\frac{1}{4}\left(ac+bc\right)\)

Tương tự cộng lại, ta có 

\(\frac{\sqrt{ab^2c^3}}{b+c}+\frac{\sqrt{bc^2a^3}}{c+a}+\frac{\sqrt{ca^2b^3}}{a+b}\le\frac{1}{2}\left(ab+bc+ca\right)\)

Mà \(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2=3\Rightarrow\frac{\sqrt{ab^2c^3}}{b+c}+...\le\frac{3}{2}\)

dấu = xảy ra <=> a=b=c=1

^.^