2 nha bạn.

2 nha bạn. mình kết bạn nha.

lm thì ko lm cứ viết đáp số thế bố cóc thèm

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

\(\Leftrightarrow ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}\ge\frac{0-1}{2}=-\frac{1}{2}\)

Dấu \(=\)khi \(\hept{\begin{cases}a+b+c=0\\a^2+b^2+c^2=1\end{cases}}\), chẳng hạn \(c=0,a=-b=\sqrt{\frac{1}{2}}\). 

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

\(< =>ab+bc+ca\le1\)

Dấu "=" tự tìm nhaaaaa

\(A=\dfrac{bc}{8a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)

\(=\dfrac{\left(bc\right)^3+8\left(ca\right)^3+8\left(ab\right)^3}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(2ca\right)^3+\left(2ab\right)^3}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(2ab+2ca\right)^3-3.2ca.2ab\left(2ab+2ca\right)}{8\left(abc\right)^2}\)

\(=\dfrac{\left(bc\right)^3+\left(-bc\right)^3-3.2ca.2ab.\left(-bc\right)}{8\left(abc\right)^2}\)

\(=\dfrac{12\left(abc\right)^2}{8\left(abc\right)^2}=\dfrac{12}{8}\)

kkkk

Thay \(c=2-\left(a+b\right)\Leftrightarrow P=2ab+c\left(a+b\right)=2ab+\left(a+b\right)\left[2-\left(a+b\right)\right]\)

\(=2ab+2\left(a+b\right)-a^2-b^2-2ab=2\left(a+b\right)-a^2-b^2=2-\left(a-1\right)^2-\left(b-1\right)^2\)

Mà \(\hept{\begin{cases}\left(a-1\right)^2\\\left(b-1\right)^2\end{cases}\ge0\forall a,b\inℝ\Rightarrow P=2-\left(a-1\right)^2-\left(b-1\right)^2\le2}\)

Dấu ''='' xảy ra \(\Leftrightarrow\) \(a=b=1\rightarrow c=0\)

Ta có: \(\dfrac{a^3+ab^2}{a^2+b+b^2}=a-\dfrac{ab}{a^2+b+b^2}\ge a-\dfrac{\sqrt[3]{a}}{3}\)

Tương tự: 

\(\Rightarrow VT\ge a+b+c-\dfrac{\Sigma\sqrt[3]{a}}{3}=3-\dfrac{\Sigma\sqrt[3]{a}}{3}\)

Áp dụng BĐT cô si chi 3 số dương, ta có:

\(a+1+1\ge3\sqrt[3]{a}\Rightarrow\dfrac{\sqrt[3]{a}}{3}\le\dfrac{a+2}{9}\)

Tương tự:

\(\Rightarrow VT\ge3-\dfrac{a+b+c+6}{9}=3-1=2\left(đpcm\right)\)

Dấu "=" xảy ra <=> a=b=c=1

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{2b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\)

\(M=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=\dfrac{x^3+y^3+z^3}{xyz}\)

\(=\dfrac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3}{xyz}=\dfrac{-z^3-3xy\left(-z\right)+z^3}{xyz}\)

\(=\dfrac{3xyz}{xyz}=3\)