Áp dụng BĐT  AM-GM ta có :

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

\(=\frac{9}{abc\left(a+b+c\right)}\ge\frac{27}{\left(ab+bc+ca\right)^2}\)

Mặt khác theo BĐT  AM-GM  có :

\(\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\le\left(\frac{a^2+b^2+c^2+2\left(ab+bc+ca\right)^3}{3}\right)=27\)

\(\Rightarrow\frac{27}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\)

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

Xét \(t+\frac{1}{t}=\frac{1}{9}+\frac{1}{t}+\frac{81}{9}.3=\frac{10}{3}\)

Vậy \(MinP=\frac{10}{3}\Leftrightarrow a=b=c=-1\)

Sửa lại chút  , vội quá nên đánh lỗi .

Xét \(t+\frac{1}{t}=\frac{1}{9}+\frac{1}{t}+\frac{8t}{9}\ge2\sqrt{\frac{t}{9}.\frac{1}{t}}+\frac{8}{9}.3=\frac{10}{3}\)

\(\Rightarrow MinP=\frac{10}{3}\Leftrightarrow a=b=c=1\)

tính hộ 1 chia 0 nha

@NTMH

Ta chứng minh \(P\ge\frac{9}{2}\). Ta đã có: \(\frac{a^3+b^3+c^3}{2abc}\ge\frac{3abc}{2abc}=\frac{3}{2}\)

Vậy cần chứng minh \(\frac{a^{2}+b^{2}}{c^{2}+ab}+\frac{b^{2}+c^{2}} {a^{2}+bc}+\frac{c^{2}+a^{2}}{b^{2}+ac}\geq 3\)

\(\Leftrightarrow a^{2}(\frac{1}{c^{2}+ab}+\frac{1}{b^{2}+ac)}+b^{2}(\frac{1}{c^{2}+ab}+\frac{1}{a^{2}+bc})+c^{2}( \frac{1}{a^{2}+bc}+\frac{1}{b^{2}+ac})\)

\(\geq \frac{4a^{2}}{(a+b)(b+c)}+\frac{4b^{2}}{(c+a)(c+b) }+\frac{4c^{2}}{(a+b)(a+c)}\)

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

BĐT đã được chứng minh

Vậy ta có \(P_{min}=\frac{9}{2}\) khi \(a=b=c\)

\(\text{⋄}\)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

\(S=\left(a^2+\frac{1}{4}\right)+\left(b^2+\frac{1}{4}\right)+\left(c^2+\frac{1}{4}\right)+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-\frac{3}{4}\)

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

\(\ge1+1+1-\frac{3}{4}=\frac{9}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{2}\)

à quên tách ra mà quên đoạn sau :v thêm vào tí nhé 

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

\(\ge2\sqrt{\frac{a}{4a}}+2\sqrt{\frac{b}{4b}}+2\sqrt{\frac{c}{4c}}+\frac{3}{4}.\frac{9}{a+b+c}-\frac{3}{4}\ge1+1+1+\frac{3}{4}.\frac{9}{\frac{3}{2}}-\frac{3}{4}=\frac{27}{4}\)

\(P=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

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

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

\(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

dấu "=" xảy ra tại a=b=c

Cách 2

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

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

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

\(\Rightarrow P\ge\frac{3}{2}\Leftrightarrow a=b=c\)

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

Đặt \(\hept{\begin{cases}b+c=x\\c+a=y\\a+b=z\end{cases}\left(x,y,z>0\right)}\)

\(\Rightarrow a=\frac{y+z-x}{2}\);\(b=\frac{z+x-y}{2}\);\(c=\frac{x+y-z}{2}\)

\(\left(1\right)\)trở thành \(\frac{y+z-x}{2x}+\frac{z+x-y}{2y}+\frac{x+y-z}{2z}\ge\frac{3}{2}\)

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

\(\Leftrightarrow\left(\frac{y}{2x}+\frac{x}{2y}\right)+\left(\frac{z}{2x}+\frac{x}{2z}\right)+\left(\frac{z}{2y}+\frac{y}{2z}\right)\ge3\)

Vì \(\frac{y}{2x}+\frac{x}{2y}\ge2\sqrt{\frac{y}{2x}.\frac{x}{2y}}=1\)( bđt AM-GM)

CMTT ​​\(\frac{z}{2x}+\frac{x}{2z}\ge1\)và \(\frac{z}{2y}+\frac{y}{2z}\ge1\)

rồi cộng vào là xong

Dấu"="xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{y}{2x}=\frac{x}{2y}\\\frac{z}{2x}=\frac{x}{2z}\\\frac{z}{2y}=\frac{y}{2z}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}2x^2=2y^2\\2z^2=2x^2\\2y^2=2z^2\end{cases}\Leftrightarrow}\hept{\begin{cases}x=y\\z=x\\y=z\end{cases}\Leftrightarrow}x=y=z\)

Vậy \(P_{min}=\frac{3}{2}\Leftrightarrow x=y=z\)

Ta có đánh giá: \(\frac{1}{2a-a^2}\ge\frac{81-108a}{25}\) \(\forall a\in\left(0;1\right)\)

Thật vậy, BĐT tương đương:

\(\left(81-108a\right)\left(2a-a^2\right)\le25\)

\(\Leftrightarrow108a^3-297a^2+162a-25\le0\)

\(\Leftrightarrow\left(3a-1\right)^2\left(25-12a\right)\ge0\) (luôn đúng \(\forall a\in\left(0;1\right)\))

Tương tự: \(\frac{1}{2b-b^2}\ge\frac{81-108b}{25}\) ; \(\frac{1}{2c-c^2}\ge\frac{81-108c}{25}\)

Cộng vế với vế:

\(\Rightarrow A\ge\frac{243-108\left(a+b+c\right)}{25}+3=\frac{42}{5}\)

\(A_{min}=\frac{42}{5}\) khi \(a=b=c=\frac{1}{3}\)