Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :

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

Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)

Do đó : \(M\ge\frac{3}{2}\)

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

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

Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)

Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

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

Áp dụng BĐT Svacxo ta có :

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

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

Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)

giải thích cho mình với, sao \(\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)  vậy

để biểu thức cho đơn giản , ta đặt x=a+1,y=b+1,z=c+1(x,y,z>0)

thì giả thiết thành \(\frac{1}{x+1}+\frac{3}{y+3}\le\frac{z}{z+2}\) .Tìm min xyz 

Áp dụng bất đẳng thức cauchy:\(\frac{z}{z+2}\ge\frac{1}{x+1}+\frac{3}{y+3}\ge2\sqrt{\frac{3}{\left(x+1\right)\left(y+3\right)}}\)(1)

từ giả thiết :\(\frac{1}{x+1}\le\frac{z}{z+2}-\frac{3}{y+3}\Leftrightarrow1-\frac{1}{x+1}\ge1-\frac{z}{z+2}+\frac{3}{y+3}\)

\(\Leftrightarrow\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\)

Áp dụng bất đẳng thức cauchy 1 lần nữa: \(\frac{x}{x+1}\ge\frac{2}{z+2}+\frac{3}{y+3}\ge2\sqrt{\frac{6}{\left(z+2\right)\left(y+3\right)}}\)(2)

tương tự ta cũng có: \(\frac{y}{y+3}\ge2\sqrt{\frac{2}{\left(z+2\right)\left(x+1\right)}}\)(3),

cả 2 vế các bất đẳng thức (1),(2)và (3) đều dương, nhân vế với vế: 

\(\frac{xyz}{\left(x+1\right)\left(y+3\right)\left(z+2\right)}\ge\frac{8.6}{\left(x+1\right)\left(z+2\right)\left(y+3\right)}\)

\(\Leftrightarrow xyz\ge48\)

Dấu = xảy ra khi x=2,y=6,z=4 hay a=1,b=5,z=3

1. Áp dụng BĐT Cauchy dạng Engle, ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

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

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

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)

\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)

Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)

\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)

Áp dụng BĐT Cauchy cho a ; b dương

Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)

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

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2-ab+1\right)\le\left(a^2+b^2-ab+1\right)\)

vì a²+b²-ab+1 >0 với mọi a,b thuộc R \(\Rightarrow\frac{1}{3}\left(a+b\right)\le1\Leftrightarrow a+b\le3\)

khi đó ta có \(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=a+\frac{4}{a}+b+\frac{1}{b}+\frac{4}{a}+\frac{1}{b}\)

\(\Leftrightarrow M=\left(a+\frac{4}{a}\right)+\left(b+\frac{1}{b}\right)+\left(\frac{4}{a}+\frac{1}{b}\right)\)

áp dụng bđt cosi cho các cặp số dương \(\left(a;\frac{4}{a}\right);\left(b;\frac{1}{b}\right);\left(\frac{4}{a};\frac{1}{b}\right)\)ta có

\(\hept{\begin{cases}a+\frac{4}{a}\ge2\cdot\sqrt{a\cdot\frac{4}{a}}=2\sqrt{4}=4\\b+\frac{1}{b}\ge2\sqrt{b\cdot\frac{1}{b}}=2\sqrt{1}=2\\\frac{4}{a}+\frac{1}{b}\ge\frac{\left(2+1\right)^2}{a+b}\ge\frac{9}{3}=3\end{cases}}\Rightarrow minM=4+3+2=9\)

\(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{1}{b}\\a=2b\end{cases}\Leftrightarrow\hept{\begin{cases}a=2\\b=1\end{cases}}}\)

vậy M đạt giá trị nhỏ nhất là 9 khi a=2; b=1

\(\text{⋄}\)Dễ có: \(B\ge\left(3+\frac{4}{a+b}\right)\left(3+\frac{4}{b+c}\right)\left(3+\frac{4}{c+a}\right)\)

\(\text{⋄}\)Đặt \(b+c=x;c+a=y;a+b=z\left(x,y,z>0\right)\)thì \(a=\frac{y+z-x}{2};b=\frac{z+x-y}{2};c=\frac{x+y-z}{2}\)

Giả thiết được viết lại thành: \(x+y+z\le3\)và ta cần tìm giá trị nhỏ nhất của \(\left(3+\frac{4}{x}\right)\left(3+\frac{4}{y}\right)\left(3+\frac{4}{z}\right)\)

\(\text{⋄}\)Ta có: \(\left(3+\frac{4}{x}\right)\left(3+\frac{4}{y}\right)\left(3+\frac{4}{z}\right)=27+36\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+48\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)+\frac{64}{xyz}\)\(\ge27+36.\frac{9}{x+y+z}+48.\frac{27}{\left(x+y+z\right)^2}+64.\frac{27}{\left(x+y+z\right)^3}\ge343\)

Đẳng thức xảy ra khi x = y = z = 1 hay a = b = c = ¹/₂

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

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

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

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

Đặt \(\left\{\begin{matrix}\frac{1}{a}=x\\\frac{1}{b}=y\\\frac{1}{c}=z\end{matrix}\right.\) suy ra \(xyz=1\). Khi đó:

\(P=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)

Áp dụng BĐT AM-GM ta có:

\(\left\{\begin{matrix}\frac{x^2}{y+z}+\frac{y+z}{4}\ge x\\\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\\\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\end{matrix}\right.\).Cộng theo vế ta có:

\(P+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow P\ge\frac{x+y+z}{2}\ge\frac{3}{2}\left(x+y+z\ge3\sqrt[3]{xyz}=3\right)\)

\(P=\left[\left(2+\frac{1}{a}+\frac{1}{b}\right)+1\right]\left[\left(2+\frac{1}{b}+\frac{1}{c}\right)+1\right]\left[\left(2+\frac{1}{c}+\frac{1}{a}\right)+1\right]\)

\(\ge\left(6\sqrt[3]{\frac{1}{4ab}}+1\right)\left(6\sqrt[3]{\frac{1}{4bc}}+1\right)\left(6\sqrt[3]{\frac{1}{4ca}}+1\right)\)

\(\ge\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ab}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4bc}}\right)^6}\right]\left[7\sqrt[7]{\left(\sqrt[3]{\frac{1}{4ca}}\right)^6}\right]\)

\(=\left[7\sqrt[7]{\left(\frac{1}{4ab}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4bc}\right)^2}\right]\left[7\sqrt[7]{\left(\frac{1}{4ca}\right)^2}\right]\)

\(=343\sqrt[7]{\left(\frac{1}{64\left(abc\right)^2}\right)^2}\ge343\sqrt[7]{\left(\frac{1}{64\left[\frac{\left(a+b+c\right)^3}{27}\right]^2}\right)^2}=343\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{2}\)

P/s: Em chưa check lại đâu nha::D

Khúc cuối bài ban nãy là \(\ge343\) nha! Em đánh nhầm

Cách khác (em thử dùng Holder, mới học nên em không chắc lắm):

\(P\ge\left(3+\sqrt[3]{\frac{1}{abc}}+\sqrt[3]{\frac{1}{abc}}\right)^3=\left(3+2\sqrt[3]{\frac{1}{abc}}\right)^3\ge\left(3+2\sqrt[3]{\frac{1}{\left[\frac{\left(a+b+c\right)^3}{27}\right]}}\right)^3\ge343\)

24+t−94(∗)
Xét hàm (∗) được: MinF(t)=F(23)=−19
⇒MinP=MinF(t)=−19.dấu "=" xảy ra khi a=b=c=13