Ta có : 

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

Áp dụng bất đẳng thức Cauchy ,ta có 

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

<=> \(2\left(a^2+b^2+c^2\right)\ge2\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

<=> \(\left(a^2+b^2+c^2\right)\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

Đẳng thức xảy ra <=> a = b = c 

Đặt \(a=x^2,b=y^3,c=z^3\)

+ \(\frac{a+b+c}{3}\ge\sqrt[3]{abc}\Leftrightarrow x^3+y^3+z^3\ge3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz\ge0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\ge0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\ge0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2+z^2-yz-xz-3xy\right)\ge0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\ge0\)

\(\Leftrightarrow\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\)

Vì BĐT cuối luôn đúng \(\forall x,y,z\ge0\) nên ta có đpcm

Dấu "=" \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

cách 2: Đặt \(a=x^3,b=y^3,c=z^3\)

Áp dụng BĐT Cauchy với 2 số ko âm :

\(\left(x^3+y^3\right)+\left(z^3+xyz\right)\ge2\sqrt{x^3y^3}+2\sqrt{xyz^4}=2\sqrt{xy}\left(xy+z^2\right)\)

Dấu "=" \(\Leftrightarrow x=y=z\)

+ \(xy+z^2\ge2\sqrt{xyz^2}=2\sqrt{xy}\cdot z\) Dấu "=" \(\Leftrightarrow z^2=xy\)

Do đó : \(x^3+y^3+z^3+xyz\ge2\sqrt{xy}\cdot2z\sqrt{xy}=4xyz\)

\(\Rightarrow x^3+y^3+z^3\ge3xyz\Rightarrow\frac{a+b+c}{3}\ge\sqrt[3]{abc}\)

Dấu "=" \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c\)

a. Đề bài sai (thực chất là nó đúng 1 cách hiển nhiên nhưng "dạng" thế này nó sai sai vì ko ai cho kiểu này cả)

Ta có: \(abc=ab+bc+ca\ge3\sqrt[3]{a^2b^2c^2}\Rightarrow abc\ge27\)

\(\Rightarrow a^2+b^2+c^2+5abc\ge a^2+b^2+c^2+5.27>>>>>8\)

b. 

\(4=ab+bc+ca+abc=ab+bc+ca+\sqrt{ab.bc.ca}\le ab+bc+ca+\sqrt{\left(\dfrac{ab+bc+ca}{3}\right)^3}\)

\(\sqrt{\dfrac{ab+bc+ca}{3}}=t\Rightarrow t^3+3t^2-4\ge0\Rightarrow\left(t-1\right)\left(t+2\right)^2\ge0\)

\(\Rightarrow t\ge1\Rightarrow ab+bc+ca\ge3\Rightarrow a+b+c\ge\sqrt{3\left(ab+bc+ca\right)}\ge3\)

- TH1: nếu \(a+b+c\ge4\)

Ta có: \(ab+bc+ca=4-abc\le4\)

\(\Rightarrow P=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)+5abc\ge4^2-2.4+0=8\)

(Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(2;2;0\right)\) và các hoán vị)

- TH2: nếu \(3\le a+b+c< 4\)

Đặt \(a+b+c=p\ge3;ab+bc+ca=q;abc=r\)

\(P=p^2-2q+5r=p^2-2q+5\left(4-q\right)=p^2-7q+20\)

Áp dụng BĐT Schur:

\(4=q+r\ge q+\dfrac{p\left(4q-p^2\right)}{9}\Leftrightarrow q\le\dfrac{p^3+36}{4p+9}\)

\(\Rightarrow P\ge p^2-\dfrac{7\left(p^3+36\right)}{4p+9}+20=\dfrac{3\left(4-p\right)\left(p-3\right)\left(p+4\right)}{4p+9}+8\ge8\)

(Dấu "=" xảy ra khi \(a=b=c=1\))

Bạn xem lại đề giùm mình nhé.

Xét: a² \(\ge\)a;   b² \(\ge\)b;  c² \(\ge\)c

\(\Rightarrow\)a² + b² + c² \(\ge\)a + b + c \(\ge\)abc

ban duong huynh giang nham roi ban oi. a²\(\ge\)a khi a\(\ge\)1 thoi. Vi du \(\frac{1}{2}^2\ge\frac{1}{2}\Rightarrow\frac{1}{4}\ge\frac{1}{2}\)(vo li)

Xét 2 trường hợp:

+ \(\left|a\right|\ge1;\left|b\right|\ge1;\left|c\right|\ge1\)

\(\Rightarrow a^2+b^2+c^2\ge a+b+c\ge abc\)

+ Trong 3 số \(\left|a\right|;\left|b\right|;\left|c\right|\)có ít nhất 1 số nhỏ hơn1

Không mất tính ttoongr quát, ta giả sử:\(\left|c\right|< 1\)

Ta có: \(a^2+b^2+c^2\ge a^2+b^2\ge2\left|ab\right|>\left|abc\right|\ge abc\)

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)