1/\(=4a^2+4b^2+c^2+8ab-4bc-4ca+4b^2+4c^2+a^2+8bc-4ca-4ab+4a^2+4c^2+b^2+8ca-4bc-4ab=\)

\(=9a^2+9b^2+9c^2=9\left(a^2+b^2+c^2\right)\)

2/

Ta có

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge-2\left(ab+bc+ca\right)=2\)

\(\Rightarrow P=9\left(a^2+b^2+c^2\right)\ge18\)

\(\Rightarrow P_{min}=18\)

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

\(\sqrt{2a^2+ab+2b^2}=\sqrt{\dfrac{3}{2}\left(a^2+b^2\right)+\dfrac{1}{2}\left(a+b\right)^2}\ge\sqrt{\dfrac{3}{4}\left(a+b\right)^2+\dfrac{1}{2}\left(a+b\right)^2}=\dfrac{\sqrt{5}}{2}\left(a+b\right)\)

Tương tự:

\(\sqrt{2b^2+bc+2c^2}\ge\dfrac{\sqrt{5}}{2}\left(b+c\right)\) ; \(\sqrt{2c^2+ca+2a^2}\ge\dfrac{\sqrt{5}}{2}\left(c+a\right)\)

Cộng vế với vế:

\(P\ge\sqrt{5}\left(a+b+c\right)\ge\dfrac{\sqrt{5}}{3}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^3=\dfrac{\sqrt{5}}{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{9}\)

Lời giải:
\((2a+2b-c)^2+(2b+2c-a)^2+(2c+2a-b)^2\)

\(=(2a+2b)^2-2c(2a+2b)+c^2+(2b+2c)^2-2a(2b+2c)+a^2+(2c+2a)^2-2b(2c+2a)+b^2\)

\(=4(a+b)^2+4(b+c)^2+4(c+a)^2+(c^2+a^2+b^2)-4c(a+b)-4b(a+c)-4a(b+c)\)

\(=4(a^2+2ab+b^2)+4(b^2+2bc+c^2)+4(c^2+2ac+a^2)+(c^2+a^2+b^2)-8(ab+bc+ac)\)

\(=9(a^2+b^2+c^2)=9.9=81\)

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 \(\hept{\begin{cases}a=b=c\\\sqrt{a}+\sqrt{b}+\sqrt{c}=3\end{cases}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}}\)

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

a) Có:

 \(a+b+c=0\\\Leftrightarrow\left(a+b+c\right)^2=0\\ \Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\\ \Leftrightarrow2ab+2bc+2ca=-1\\ \Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\\ \Leftrightarrow\left(ab+bc+ca\right)^2=\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\\ \Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\dfrac{1}{4}-0=\dfrac{1}{4} \)

câu (b) cho đa thức P (x) = cái gì?

mong các thầy cô giúp em giải bài này với ạ