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

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

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

\(\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\Leftrightarrow b-2\sqrt{bc}+c\ge0\Leftrightarrow b+c\ge2\sqrt{bc}\) dấu "="xảy ra khi b=c

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

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

tương tự ta có \(\hept{\begin{cases}\sqrt{\left(b+2c\right)\left(b+2c\right)}\ge b+2\sqrt{bc}\\\sqrt{\left(c+2a\right)\left(a+2b\right)}\ge c+2\sqrt{ab}\end{cases}}\)

dấu "=" xảy ra khi a=b=c

\(\Rightarrow A=\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2a\right)\left(c+2b\right)}\)\(\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)

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

dấu "="xảy ra khi \(\hept{\begin{cases}a=b=c\\\sqrt{a}+\sqrt{b}+\sqrt{c}=3\end{cases}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}}\)

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

làm ơn trả lời hộ mk với ah mai mk phải nộp bài r

BĐT cần  chứng minh tương đương với :

\(\left(a^2b+b^2c+c^2a\right)\left(2+\frac{1}{a^2b^2c^2}\right)\ge9\)

\(\Leftrightarrow2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge9\)

Áp dụng BĐT Cô-si cho 3 số dương ,ta có :

\(a^2b+a^2b+\frac{1}{ab^2}\ge3\sqrt[3]{a^2b.a^2b.\frac{1}{ab^2}}=3a\)

tương tự :  \(b^2c+bc^2+\frac{1}{bc^2}\ge3b\), \(\left(c^2a+ca^2+\frac{1}{ca^2}\right)\ge3c\)

Cộng 3 BĐT trên theo vế, ta được :

\(2\left(a^2b+b^2c+c^2a\right)+\frac{1}{ab^2}+\frac{1}{bc^2}+\frac{1}{ca^2}\ge3\left(a+b+c\right)=9\)

Dấu "=" xảy ra khi a = b = c = 1

Sai đề! Sửa: that 2c+b-a=2c+a-b

Đặt 2a+b-c=x, 2b+c-a=y, 2c+a-b=z

\(\Rightarrow8\left(a+b+c\right)^3=\left(x+y+z\right)^3=x^3+y^3+z^3\)và \(P=\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

Ta có: \(\left(x+y+z\right)^3-x^3-y^3-z^3=0\Leftrightarrow\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)-x^3-y^3=0\)

\(\Leftrightarrow3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)=0\Leftrightarrow3\left(x+y\right)\left(xy+xz+yz+z^2\right)=0\)

\(\Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\Leftrightarrow3P=0\Leftrightarrow P=0\)

\(\dfrac{a}{a+2b^3}=a-\dfrac{2ab^3}{a+b^3+b^3}\ge a-\dfrac{2ab^3}{3\sqrt[3]{ab^6}}=a-\dfrac{2}{3}.b\sqrt[3]{a^2}\ge a-\dfrac{2}{9}b\left(a+a+1\right)\)

\(\Rightarrow\dfrac{a}{a+2b^3}\ge a-\dfrac{2}{9}\left(2ab+b\right)\)

Tương tự: \(\dfrac{b}{b+2c^3}\ge b-\dfrac{2}{9}\left(2bc+c\right)\) ; \(\dfrac{c}{c+2a^3}\ge c-\dfrac{2}{9}\left(2ac+a\right)\)

Cộng vế:

\(A\ge a+b+c-\dfrac{2}{9}\left(2ab+2bc+2ca+a+b+c\right)=3-\dfrac{2}{9}\left[2\left(ab+bc+ca\right)+3\right]\)

\(A\ge3-\dfrac{2}{9}\left[\dfrac{2}{3}\left(a+b+c\right)^2+3\right]=1\)