Ta có \(a^3+b^3+c^3-3abc=a^3+b^3+c^3-abc-abc-abc+ac^2+a^2c-ac^2-a^2c+ab^2+a^2b-ab^2-a^2b+b^2c+bc^2-b^2c-bc^2=a^2\left(a+b+c\right)+b^2\left(a+b+c\right)+c^2\left(a+b+c\right)-ab\left(a+b+c\right)-bc\left(a+b+c\right)-ac\left(a+b+c\right)=\frac{1}{2}\left(a+b+c\right)\left(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right)\)=0 vạy a+b+c=0

Lời giải:

Áp dụng BĐT Cô-si: 

$a+b+c\geq 3\sqrt[3]{abc}=3(1)$
Tiếp tục áp dụng BĐT Cô-si:

$a^3+a\geq 2a^2$

$b^3+b\geq 2b^2$

$c^3+c\geq 2c^2$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)$

Lại có:

$a^2+1\geq 2a$

$b^2+1\geq 2b$

$c^2+1\geq 2c$

$\Rightarrow a^2+b^2+c^2\geq 2(a+b+c)-3=(a+b+c)+(a+b+c)-3$

$\geq a+b+c+3-3=a+b+c(2)$

$\Rightarrow a^3+b^3+c^3\geq 2(a^2+b^2+c^2)-(a+b+c)\geq a^2+b^2+c^2(3)$

Từ $(1); (2); (3)$ ta có đpcm.

ta có \(a^3+b^3+a^2c+b^2c-abc=\left(a+b\right)\left(a^2-ab+b^2\right)+c\left(a^2-ab+b^2\right)=\left(a^2-ab+b^2\right)\left(a+b+c\right)\)

mà a+b+c=0

\(\Rightarrow a^3+b^3+a^2c+b^2c-abc=\left(a^2-ab+b^2\right).0=0\left(đpcm\right)\)

Ta có: \(x^3+y^{ 3}=\left(x+y\right)\left(x^2-xy+y^2\right)\ge\left(x+y\right)\left(2xy-xy\right)=\left(x+y\right)xy,\forall x,y\ge0\)

Áp dụng:

\(\sum_{cyc}\dfrac{1}{a^3+b^3+abc}\le\sum_{cyc}\dfrac{1}{\left(a+b\right)ab+abc}=\sum_{cyc}\dfrac{1}{ab\left(a+b+c\right)}=\dfrac{a+b+c}{abc\left(a+b+c\right)}=\dfrac{1}{abc}\)

\("="\Leftrightarrow a=b=c\)

Ta có: \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)\ge\left(x+y\right)\left(2xy-xy\right)=\left(x+y\right)xy\)( \(\forall x,y\ge0\) )

Áp dụng: \(\sum\dfrac{1}{a^3+b^3+abc}\le\dfrac{1}{\left(a+b\right)ab+abc}=\sum\dfrac{1}{ab\left(a+b+c\right)}=\dfrac{a+b+c}{abc\left(a+b+c\right)}=\dfrac{1}{abc}\)

\("="\Leftrightarrow a=b=c\)

ta chứng minh đc \(x^3+y^3\ge xy\left(x+y\right)\)

thay vào + biến đổi ta có đpcm

đẳng thúc xảy ra khi a=b=c

lol!!!

3: \(\left\{{}\begin{matrix}a+b>=2\sqrt{ab}\\b+c>=2\sqrt{bc}\\a+c>=2\sqrt{ac}\end{matrix}\right.\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)>=8abc\)

1: =>(a+b)(a^2-ab+b^2)-ab(a+b)>=0

=>(a+b)(a^2-2ab+b^2)>=0

=>(a+b)(a-b)^2>=0(luôn đúng)

2) Áp dụng bất đẳng thức ở câu 1 ta có:

\(\dfrac{1}{a^3+b^3+abc}\le\dfrac{1}{ab\left(a+b\right)+abc}=\dfrac{1}{ab\left(a+b+c\right)}\)

Tương tự: \(\dfrac{1}{b^3+c^3+abc}\le\dfrac{1}{bc\left(a+b+c\right)}\)

và \(\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{ca\left(a+b+c\right)}\)

Cộng vế theo vế của các bất đẳng thức trên ta được:

\(\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\left(đpcm\right)\)

Dấu "=" xảy ra khi a=b=c.

Ta có BĐT:

\(a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)\)\(\ge\left(a+b\right)ab\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Rightarrow\dfrac{1}{a^3+b^3+abc}\le\dfrac{1}{ab\left(a+b+c\right)}\)

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT\le\dfrac{1}{ab\left(a+b+c\right)}+\dfrac{1}{bc\left(a+b+c\right)}+\dfrac{1}{ac\left(a+b+c\right)}\)

\(=\dfrac{a+b+c}{abc\left(a+b+c\right)}=\dfrac{1}{abc}=VP\)

Khi \(a=b=c\)