Đặt \(P=\dfrac{a^3}{a^2+b^2+ab}+\dfrac{b^3}{b^2+c^2+bc}+\dfrac{c^3}{c^2+a^2+ca}\)

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

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

Cộng vế:

\(P\ge\dfrac{a+b+c}{3}=673\)

Dấu "=" xảy ra khi \(a=b=c=673\)

Ta có $(a^2+2)(b^2+2)(c^2+2)$

$=a^2b^2c^2+2\sum a^2b^2+4(a^2+b^2+c^2)+8

Suy ra $(a^2+2)(b^2+2)(c^2+2)-18-3(a^2+b^2+c^2)$

$=a^2b^2c^2+2\sum a^2b^2+(a^2+b^2+c^2)-10$

Đặt $s=a^2+b^2+c^2$

Ta có $s\ge ab+bc+ca=3$ và $\sum a^2b^2\ge ab+bc+ca=3$

(vì $ab+bc+ca=3$ và $\sum a^2b^2\ge \dfrac{(ab+bc+ca)^2}{3}$).

Do đó $a^2b^2c^2+2\sum a^2b^2+s-10$$\ge 0+2\cdot3+3-10$$=-1$

Mặt khác $\sum a^2b^2\ge \dfrac{(ab+bc+ca)^2}{3}=3$ nên $a^2b^2c^2+2\sum a^2b^2+s-10$

$\ge a^2b^2c^2+s-4$

Lại có $(ab+bc+ca)^2\ge 3abc(a+b+c)$

$\Rightarrow a+b+c\le \dfrac3{abc}$ và $s=(a+b+c)^2-6\ge \dfrac9{a^2b^2c^2}-6$

Đặt $t=abc$.

Khi đó $a^2b^2c^2+s-4\ge t^2+\dfrac9{t^2}-10$

$=\left(t-\dfrac3t\right)^2-4\ge0$

(vì từ $ab+bc+ca=3$ suy ra $t\le1$).

Vậy $\boxed{(a^2+2)(b^2+2)(c^2+2)-18\ge 3(a^2+b^2+c^2)}.$

Dấu bằng khi $\boxed{a=b=c=1.}$

Xin lỗi nhé!

Áp dụng BĐT ta có:
`a^2+9>=6a`
`b^2+25>=10b`
`c^2+4>=4a`
`=>a^2+b^2+c^2+38>=6a+10b+4c`
`<=>76>=6a+10b+4c(1)`
Ta có:
`6a+10b+4c`
`=6(a+b)+4(b+c)`
`=48+4(b+c)>=48+4.7=76(2)`
`(1)(2)=>6a+10b+4c=76`
`<=>a=3,b=5,c=2`

Do \(a^2+b^2+c^2=38\Rightarrow\left|b\right|\le\sqrt{38}< 7\)

\(\Rightarrow c\ge7-b>0\)

\(\Rightarrow c^2\ge\left(7-b\right)^2\)

Do đó:

\(38=\left(8-b\right)^2+b^2+c^2\ge\left(8-b\right)^2+b^2+\left(7-b\right)^2\)

\(\Leftrightarrow5\left(b-5\right)^2\le0\)

\(\Leftrightarrow b=5\Rightarrow a=3;c=2\)

a)

$A=\dfrac{a}{b^2+1}+\dfrac{b}{c^2+1}+\dfrac{c}{a^2+1}$

$\ge \dfrac{(a+b+c)^2}{a(b^2+1)+b(c^2+1)+c(a^2+1)}\qquad (\text{Cauchy Engel})$

$=\dfrac{1}{ab^2+bc^2+ca^2+1}$

$\ge \dfrac{1}{ab(a+b)+bc(b+c)+ca(c+a)+1}$

$=\dfrac{1}{(a+b+c)(ab+bc+ca)+1}

$\ge \dfrac{1}{\frac13+1}$ $=\dfrac34$

Dấu bằng khi $a=b=c=\dfrac13$.

$\boxed{A_{\min}=\dfrac34}$

b)

$B=\dfrac{a}{ab+2c}+\dfrac{b}{bc+2a}+\dfrac{c}{ca+2b}$

$\ge \dfrac{(a+b+c)^2}{a(ab+2c)+b(bc+2a)+c(ca+2b)}$

$=\dfrac4{a^2b+b^2c+c^2a+2(ab+bc+ca)}$

Lại có $a^2b+b^2c+c^2a\le (a+b+c)(ab+bc+ca)$$=2(ab+bc+ca)$

Nên $B\ge \dfrac4{4(ab+bc+ca)}$$=\dfrac1{ab+bc+ca}$

$\ge \dfrac1{\frac{(a+b+c)^2}{3}}$ $=\dfrac34$

Dấu bằng khi $a=b=c=\dfrac23$.

$B_{\min}=\dfrac34$

Ta có:

 \(a^2+b^2+c^2=ab+bc+ca\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\ \Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

Mà \(\left(a-b\right)^2,\left(b-c\right)^2,\left(c-a\right)^2\ge0\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Rightarrow\left(a-b\right)^2=\left(b-c\right)^2=\left(c-a\right)^2=0\\ \Leftrightarrow a=b=c\)

Lại có: \(a+b+c=3\Rightarrow a=b=c=1\)

\(\Rightarrow M=1^{2016}+1^{2015}+1^{2020}=1+1+1=3\)

Đề là tìm GTNN hay GTLN hả bạn?

Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)

\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ; \(\forall a;b;c\)

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

\(\Rightarrow ab+bc+ca\le1\)

\(\Rightarrow P_{max}=1\) khi \(a=b=c\)

Lại có:

\(\left(a+b+c\right)^2\ge0\) ; \(\forall a;b;c\)

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

\(\Leftrightarrow ab+bc+ca\ge-\dfrac{a^2+b^2+c^2}{2}=-\dfrac{1}{2}\)

\(P_{min}=-\dfrac{1}{2}\) khi \(a+b+c=0\)