Áp dụng BĐT Svacxo ta có :

\(\frac{1}{a^3\left(7b+3c\right)}+\frac{1}{b^3\left(7c+3a\right)}+\frac{1}{c^3\left(7a+3b\right)}=\frac{\frac{1}{a^2}}{7ab+7ac}+\frac{\frac{1}{b^2}}{7bc+3ab}+\frac{\frac{1}{c^2}}{7ac+3bc}\)

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{10\left(ab+bc+ca\right)}=\frac{1}{10}.\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{ab+bc+ca}=\frac{1}{10}.\left(ab+bc+ca\right)\)

\(=\frac{1}{10}.\frac{ab+bc+ca}{abc}=\frac{1}{10}.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(đpcm\right)\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)thì \(x,y,z>0\)và ta cần chứng minh \(\frac{x}{\sqrt{3zx+yz}}+\frac{y}{\sqrt{3xy+zx}}+\frac{z}{\sqrt{3yz+xy}}\ge\frac{3}{2}\)\(\Leftrightarrow\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\frac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng phân thức, ta có: \(\frac{x^2}{x\sqrt{3zx+yz}}+\frac{y^2}{y\sqrt{3xy+zx}}+\frac{z^2}{z\sqrt{3yz+xy}}\ge\)\(\frac{\left(x+y+z\right)^2}{x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}}\)

Áp dụng BĐT Cauchy-Schwarz, ta có: \(x\sqrt{3zx+yz}+y\sqrt{3xy+zx}+z\sqrt{3yz+xy}\)\(=\sqrt{x}.\sqrt{3zx^2+xyz}+\sqrt{y}.\sqrt{3xy^2+xyz}+\sqrt{y}.\sqrt{3yz^2+xyz}\)\(\le\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\)

Ta cần chứng minh \(\sqrt{\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]}\le\frac{2}{3}\left(x+y+z\right)^2\)

\(\Leftrightarrow\left(x+y+z\right)^4\ge\frac{9}{4}\left(x+y+z\right)\left[3\left(xy^2+yz^2+zx^2+xyz\right)\right]\)

\(\Leftrightarrow\left(x+y+z\right)^3\ge\frac{27}{4}\left(xy^2+yz^2+zx^2+xyz\right)\)(*)

Không mất tính tổng quát, giả sử \(y=mid\left\{x,y,z\right\}\)thì khi đó \(\left(y-x\right)\left(y-z\right)\le0\Leftrightarrow y^2+zx\le xy+yz\)

\(\Leftrightarrow xy^2+zx^2\le x^2y+xyz\Leftrightarrow xy^2+yz^2+zx^2+xyz\le\)\(x^2y+yz^2+2xyz=y\left(z+x\right)^2=4y.\frac{z+x}{2}.\frac{z+x}{2}\)

\(\le\frac{4}{27}\left(y+\frac{z+x}{2}+\frac{z+x}{2}\right)^3=\frac{4\left(x+y+z\right)^3}{27}\)

Như vậy (*) đúng

Đẳng thức xảy ra khi a = b = c

\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)

\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)

\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)

\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)

\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

Vậy VT = VP, đẳng thức được chứng minh

bài 2 thì bạn áp dụng bdt cô si với lựa chọn điểm rơi  hoặc bdt holder  ( nó giống kiểu bunhia ngược ) . bai 1 thi ap dung cai nay \(\frac{1}{x}+\frac{1}{y}>=\frac{1}{x+y}\)  câu 1 khó hơn nhưng bạn biết lựa chọn điểm rơi với áp dụng bdt phụ kia là ok .

Bài 1:Đặt VT=A

Dùng BĐT \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{x+y+z}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)x,y,z>0\)

Áp dụng vào bài toán trên với x=a+c;y=b+a;z=2b ta có:

\(\frac{ab}{a+3b+2c}=\frac{ab}{\left(a+c\right)+\left(b+c\right)+2b}\le\frac{ab}{9}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{2b}\right)\)

Tương tự với 2 cái còn lại

\(A\le\frac{1}{9}\left(\frac{bc+ac}{a+b}+\frac{bc+ab}{a+c}+\frac{ab+ac}{b+c}\right)+\frac{1}{18}\left(a+b+c\right)\)

\(\Rightarrow A\le\frac{1}{9}\left(a+b+c\right)+\frac{1}{18}\left(a+b+c\right)=\frac{a+b+c}{6}\)

Đẳng thức xảy ra khi a=b=c 

Bài 2:

Biến đổi BPT \(4\left(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}\right)\ge3\)

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

Dự đoán điểm rơi xảy ra khi a=b=c=1

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge\frac{3a}{4}\)

Tương tự suy ra

\(VT\ge\frac{2\left(a+b+c\right)-3}{4}\ge\frac{2\cdot3\sqrt{abc}-3}{4}=\frac{3}{4}\)

Áp dụng BĐT \(\sqrt{xy}\le\frac{x+y}{2}\)

\(VT=\frac{2\left(a+b+c\right)}{\sqrt{4a\left(a+3b\right)}+\sqrt{4b\left(b+3c\right)}+\sqrt{4c\left(c+3a\right)}}\)

\(\Rightarrow VT\ge\frac{2\left(a+b+c\right)}{\frac{4a+a+3b}{2}+\frac{4b+b+3c}{2}+\frac{4c+c+3a}{2}}\)

\(\Rightarrow VT\ge\frac{4\left(a+b+c\right)}{8\left(a+b+c\right)}=\frac{1}{2}\) (đpcm)

Dấu "=" khi \(a=b=c\)

\(P\ge\frac{\left(a+b+b+c+c+a\right)^2}{b+3c+c+3a+a+3b}=\frac{4\left(a+b+c\right)^2}{4\left(a+b+c\right)}=a+b+c\)

Dấu "=" xảy ra khi \(a=b=c\)