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!!!

Lời giải:
Trước tiên ta đi cm bất đẳng thức sau: với \(a,b>0\) thì \(a^3+b^3\geq ab(a+b)\)

BĐT đúng vì nó tương đương với \((a-b)^2(a+b)\geq 0\) ( luôn đúng)

Do đó:, kết hợp với \(abc=1\Rightarrow \)\(\frac{1}{a^3+b^3+abc}\leq \frac{1}{ab(a+b+c)}=\frac{c}{a+b+c}\)

Tương tự với các phân thức còn lại và cộng theo vế:

\(\Rightarrow \text{VT}\leq \frac{a+b+c}{a+b+c}=1=\frac{1}{abc}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Có: \(\left(a-b\right)^2\ge0\Rightarrow\left(a-b\right)^2.\left(a+b\right)\ge0\Leftrightarrow a^3+b^3-ab\left(a+b\right)\ge0\)

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

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

TT: \(\frac{1}{b^3+c^3+abc}\le\frac{1}{bc\left(a+b+c\right)}\)

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

Cộng vế với vế ta được:

\(\frac{1}{a^3+b^3+abc}+\frac{1}{b^3+c^3+abc}+\frac{1}{c^3+a^3+abc}\le\frac{1}{a+b+c}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\le\frac{1}{a+b+c}.\frac{c+a+b}{abc}=\frac{1}{abc}\left(đpcm\right)\)

Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm

\(\Rightarrow\left\{\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ca\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a^2-ab+b^2\ge ab\\b^2-bc+c^2\ge bc\\c^2-ca+a^2\ge ca\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\\\left(b+c\right)\left(b^2-bc+c^2\right)\ge bc\left(b+c\right)\\\left(c+a\right)\left(c^2-ca+a^2\right)\ge ca\left(c+a\right)\end{matrix}\right.\)

Áp dụng hẳng đẳng thức tổng 2 lập phương

\(\Rightarrow\left\{\begin{matrix}a^3+b^3\ge ab\left(a+b\right)\\b^3+c^3\ge bc\left(b+c\right)\\c^3+a^3\ge ca\left(c+a\right)\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a^3+b^3+abc\ge ab\left(a+b\right)+abc\\b^3+c^3+abc\ge bc\left(b+c\right)+abc\\c^3+a^3+abc\ge ca\left(c+a\right)+abc\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}a^3+b^3+abc\ge ab\left(a+b+c\right)\\b^3+c^3+abc\ge bc\left(a+b+c\right)\\c^3+a^3+abc\ge ca\left(a+b+c\right)\end{matrix}\right.\)

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

\(\Rightarrow VT\le\dfrac{abc}{ab\left(a+b+c\right)}+\dfrac{abc}{bc\left(a+b+c\right)}+\dfrac{abc}{ca\left(a+b+c\right)}\)

\(\Rightarrow VT\le\dfrac{c}{a+b+c}+\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}\)

\(\Rightarrow VT\le\dfrac{a+b+c}{a+b+c}=1\)

\(\Leftrightarrow VT\le\dfrac{1}{abc}\)

\(\Leftrightarrow\dfrac{1}{a^3+b^3+abc}+\dfrac{1}{b^3+c^3+abc}+\dfrac{1}{c^3+a^3+abc}\le\dfrac{1}{abc}\)

\(\Rightarrow\) ( đpcm )

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\text{≥}\) \(\left(a+b\right)ab\)

⇒ \(a^3+b^3+abc\text{≥}\left(a+b\right)ab+abc=ab\left(a+b+c\right)\)

Tương tự : \(b^3+c^3+abc\text{ ≥}\left(b+c\right)bc+abc=bc\left(a+b+c\right)\)

\(c^3+a^3+abc\text{ ≥}\left(a+c\right)ac+abc=ac\left(a+b+c\right)\)

⇒ \(VT\text{ }\text{≤}\dfrac{1}{a+b+c}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)=\dfrac{1}{a+b+c}.\dfrac{a+b+c}{abc}=\dfrac{1}{abc}\)

Tham khảo nhá

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.

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\) với \(x;y;z>0\Rightarrow xyz=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có: \(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(P=\dfrac{1}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}+\dfrac{1}{\left(y^2+z^2\right)+\left(z^2+1\right)+2}+\dfrac{1}{\left(z^2+x^2\right)+\left(x^2+1\right)+2}\)

\(P\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{xyz}{yz+z+xyz}+\dfrac{y}{xyz+xy+y}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{xy}{y+1+xy}+\dfrac{y}{1+xy+y}\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)