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

Dấu "=" xảy ra khi \(a=b=c\)

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)+abc\)

\(=abc+a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+abc+abc\)

\(=\left(a+b+c\right)\left(ab+bc+ca\right)\)( phân tích nhân tử các kiểu )

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge\left(a+b+c\right)\left(ab+bc+ca\right)-abc\left(1\right)\)

\(a+b+c\ge3\sqrt[3]{abc};ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow\left(a+b+c\right)\left(ab+bc+ca\right)\ge9abc\)

\(\Rightarrow-abc\ge\frac{-\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

Khi đó:\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(\ge\left(a+b+c\right)\left(ab+bc+ca\right)-\frac{\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\)

\(=\frac{8\left(a+b+c\right)\left(ab+bc+ca\right)}{9}\left(2\right)\)

Từ ( 1 ) và ( 2 ) có đpcm

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 \(a=b=c\)

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

ta có:

\(\left(b-c\right)^2\ge0\Leftrightarrow b^2+4bc+4c^2\le3b^2+6c^2\Leftrightarrow\left(b+2c\right)^2\le3b^2+6c^2\)

\(\Leftrightarrow\frac{\left(b+2c\right)^2}{3b^2+6c^2}\le1\Leftrightarrow\frac{b+2c}{\sqrt{3b^2+6c^2}}\le1\Leftrightarrow\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}\le a\)

cmtt =>\(\frac{a\left(b+2c\right)}{\sqrt{3b^2+6c^2}}+\frac{b\left(c+2a\right)}{\sqrt{3c^2+6a^2}}+\frac{c\left(a+2b\right)}{\sqrt{3a^2+6b^2}}\le a+b+c\left(Q.E.D\right)\)

dấu = xảy ra khi a=b=c

Xí trước phần b

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

\(=\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(c+a\right)}+\frac{abc}{c^3\left(a+b\right)}\)

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

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

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

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)

Cách làm khác của phần b ngắn gọn hơn:)

Ta có; \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\)

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

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

\(\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(\frac{ab+bc+ca}{abc}\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\ge\frac{3\sqrt[3]{\left(abc\right)^2}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

Phần a không thể CM toàn bộ bằng BĐT rồi, bắt buộc vẫn phải sử dụng biến đổi tương đương

Ta có: \(\frac{1}{a+3b}+\frac{1}{b+3c}+\frac{1}{c+3a}\ge\frac{\left(1+1+1\right)^2}{a+3b+b+3c+c+3a}=\frac{9}{4\left(a+b+c\right)}\)

Bây giờ ta cần CM: \(\frac{9}{4\left(a+b+c\right)}\ge\frac{3}{3+abc}\)\(\left(0\right)\)

\(\Leftrightarrow9\left(3+abc\right)\ge12\left(a+b+c\right)\)

\(\Leftrightarrow9+3abc\ge4\left(a+b+c\right)\)

Đặt \(\hept{\begin{cases}a=1-x\\b=1-y\\c=1-z\end{cases}}\Rightarrow\left(x,y,z\right)\in\left[0,1\right]\)

Thay vào ta được: \(9+3\left(1-x\right)\left(1-y\right)\left(1-z\right)\ge4\left(3-x-y-z\right)\)

\(\Leftrightarrow9+3-3\left(x+y+z\right)+3\left(xy+yz+zx\right)-3xyz\ge12-4\left(x+y+z\right)\)

\(\Leftrightarrow x+y+z+3\left(xy+yz+zx\right)-3xyz\ge0\) \(\left(1\right)\)

Lại có: \(\hept{\begin{cases}x+y+z\ge3\sqrt[3]{xyz}\ge3xyz\\3\left(xy+yz+zx\right)\ge3\sqrt[3]{\left(xyz\right)^2}\ge9xyz\end{cases}}\) vì \(\left(x,y,z\right)\in\left[0,1\right]\)

\(\left(1\right)\ge3xyz+9xyz-3xyz=9xyz\ge0\left(\forall x,y,z\right)\)

=> (1) luôn đúng

=> (0) luôn đúng

=> đpcm

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

\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

đây\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)