ta có : \(a^8+b^8-a^6b^2-a^2b^6\ne\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\)

và \(a^2b^2\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\) cũng có thể âm

\(\Rightarrow\) sai

bạn chép lại đề nha

=a³+a^2c+a^2b-a^2b-abc+b^2c+b^3+b^2a-b^2a

=a^2(a+b+c)-a^2b-abc+b^2(a+b+c)-b^2a

=     -a^2b-abc-b^2a

=     -ab(a+b+c)=-ab 0 =0

vậy đa thức này bằng 0

dfgdfg

Bài 2:

Áp dụng BĐT: \(x^2+y^2+z^2\ge xy+yz+xz\), ta có:

\(a^4+b^4+c^4\ge a^2b^2+b^2c^2+a^2c^2\) (1)

Lại áp dụng tương tự ta có:

\(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge ab^2c+abc^2+a^2bc\)

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

Từ (1) và (2) suy ra:

\(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

Bài 1:

Áp dụng BĐT Cô -si, ta có:

\(\dfrac{a^2}{b^3}+\dfrac{1}{a}+\dfrac{1}{a}\ge\sqrt[3]{\dfrac{a^2}{b^3}.\dfrac{1}{a}.\dfrac{1}{a}}=\dfrac{3}{b}\)

\(\dfrac{b^2}{c^3}+\dfrac{1}{b}+\dfrac{1}{b}\ge\sqrt[3]{\dfrac{b^2}{c^3}.\dfrac{1}{b}.\dfrac{1}{b}}=\dfrac{3}{c}\)

\(\dfrac{c^2}{a^3}+\dfrac{1}{c}+\dfrac{1}{c}\ge\sqrt[3]{\dfrac{c^2}{a^3}.\dfrac{1}{c}.\dfrac{1}{c}}=\dfrac{3}{a}\)

Cộng vế theo vế ta được:

\(\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{a^2}{a^3}+\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\ge3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{a^2}{b^3}+\dfrac{b^2}{c^3}+\dfrac{c^2}{a^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

p/s: không chắc lắm, có gì sai xót xin giúp đỡ

\(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)

\(a^3+b^3+a^2c+b^2c\)

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

\(=-ba^2-ab^2\)

\(=-ab\left(a+b\right)\)

\(=-ab\cdot\left(-c\right)\)

\(=abc\) (đpcm)

   Ta có   a + b + c + d = 0

\(\Leftrightarrow\)a+c = -( b+ d)

\(\Leftrightarrow\)(a+c)³ = - ( b+d)³ 

\(\Leftrightarrow\)a³ + c³ + 3ac.(a+c) = - [ b³ + d³ + 3bd( b+d) ]

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = -3bd(b+d) - 3ac(a+c)

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = -3bd( b+d) + 3ac( b+d)   

\(\Leftrightarrow\)a³ + b³ + c³ + d³ = 3( ac - bd)(b +d) (đpcm)

Ta có:     a + b + c +d = 0 => a + b + (c+d) = 0

=> a³ + b³ +(c+d)³ = 3ab(c+d)

=>a³ +b³ +c³ +d³ +3cd(c+d) = 3ab(c+d)

=> a³ +b³ +c³ +d³  = 3ab(c+d) – 3cd(c+d) = 3(c+d)(ab – cd).

Tớ chưa học bđt Cauchy-Schwwarz và hệ quả AM-GM thì sao?

\(\dfrac{a^3}{b+c}+\dfrac{b^3}{a+c}+\dfrac{c^3}{a+b}\)

\(=\dfrac{a^4}{ab+ac}+\dfrac{b^4}{ab+bc}+\dfrac{c^4}{ac+bc}\)

\(\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2+b^2+c^2\right)}\)

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

Dấu "=" xảy ra khi: \(a=b=c=\dfrac{1}{\sqrt{3}}\)