\(\frac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}=\frac{a^4}{\left[\left(a-b\right)\left(a+b\right)+c^2\right]\left[\left(a-c\right)\left(a+c\right)+b^2\right]}\)

\(\frac{a^4}{\left[-c\left(a-b\right)+c^2\right]\left[-b\left(a-c\right)+b^2\right]}=\frac{a^4}{4bc\left(b+c\right)^2}=\frac{a^4}{4a^2bc}\)

Tương tự với 2 phân thức còn lại, ta cũng có : \(\frac{b^4}{b^4-\left(c^2-a^2\right)^2}=\frac{b^4}{4ab^2c};\frac{c^4}{c^4-\left(a^2-b^2\right)^2}=\frac{c^4}{4abc^2}\)

\(VT=\frac{a^4}{4a^2bc}+\frac{b^4}{4ab^2c}+\frac{c^4}{4abc^2}=\frac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}=\frac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\)

\(VT=\frac{a^3+b^3+c^3}{4abc}\)

Mà \(a+b+c=0\) nên \(a^3+b^3+c^3=3abc\) ( tự cm ) 

\(\Rightarrow\)\(VT=\frac{3abc}{4abc}=\frac{3}{4}\) ( đpcm ) 

Chúc bạn học tốt ~ 

(a^2+b^2+c^2) x 2 = 2 x (a^4+b^4+c^4)

suy ra: (a+b+c)^2 x 2 = (a+b+c)^4 x 2

Mà a+b+c= 0(gt)

suy ra: 0^2 x 2=0^4 x 2

0 = 0

=)))

a+b+c=0⇔a3+b3+c3=3abca+b+c=0⇔a3+b3+c3=3abc (cái này tự chứng minh nhá, dễ)

⇒3abc(a2+b2+c2)=(a3+b3+c3)(a2+b2+c2)=a5+b5+c5+a3(b2+c2)+b3(c2+a2)+c3(a2+b2)⇒3abc(a2+b2+c2)=(a3+b3+c3)(a2+b2+c2)=a5+b5+c5+a3(b2+c2)+b3(c2+a2)+c3(a2+b2)

Lại có b+c=−a⇔b2+c2=(b+c)2−2bc=a2−2bcb+c=−a⇔b2+c2=(b+c)2−2bc=a2−2bc

Tương tự c2+a2=b2−2ac,a2+b2=c2−2abc2+a2=b2−2ac,a2+b2=c2−2ab

Nên 3abc(a2+b2+c2)=a5+b5+c5+a3(a2−2bc)+b3(b2−2ac)+c3(c2−2ab)=2(a5+b5+c5)−2abc(a2+b2+c2)

Ta có

\(\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)=a^5+a^2b^3+a^2c^3+a^3b^2+b^5+b^2c^3+a^3c^2+b^3c^2+c^5\)

\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)-a^2b^2\left(a+b\right)-b^2c^2\left(b+c\right)-a^2c^2\left(a+c\right)\)

Do a+b+c=0

=> a+b=-c; b+c=-a; a+c=-b

\(\Rightarrow a^5+b^5+c^5=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+a^2b^2c+ab^2c^2+a^2bc^2=\)

\(=\left(a^2+b^2+c^2\right)\left(a^3+b^3+c^3\right)+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right)\left[\left(a+b\right)^3-3ab\left(a+b\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right).\left[\left(-c^3\right)-3ab.\left(-c\right)+c^3\right]+abc\left(ab+bc+ac\right)=\)

\(=\left(a^2+b^2+c^2\right).3abc+abc\left(ab+bc+ab\right)=\)

\(=abc.\left[3\left(a^2+b^2+c^2\right)+ab+bc+ac\right]=\)

\(=abc\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{a^2+b^2+c^2+2ab+2bc+2ac}{2}\right]=\)

\(=abc.\left[\dfrac{5}{2}.\left(a^2+b^2+c^2\right)+\dfrac{\left(a+b+c\right)^2}{2}\right]=\)

\(=abc.\dfrac{5}{2}.\left(a^2+b^2+c^2\right)\)

\(\Rightarrow\dfrac{a^5+b^5+c^5}{5}=abc.\dfrac{a^2+b^2+c^2}{2}\left(đpcm\right)\)

Đặt :

\(A=\)\(\dfrac{a^4}{a^4-\left(b^2-c^2\right)^2}+\dfrac{b^4}{b^4-\left(c^2-a^2\right)^2}+\dfrac{c^4}{c^4-\left(a^2-b^2\right)}\)

\(=\dfrac{a^4}{\left(a^2-b^2+c^2\right)\left(a^2+b^2-c^2\right)}+\dfrac{b^4}{\left(b^2-c^2+a^2\right)\left(b^2+c^2-a^2\right)}+\dfrac{c^4}{\left(c^2-a^2+b^2\right)\left(c^2+a^2-b^2\right)}\)

Ta có : \(a+b+c=0\)

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

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

\(\Leftrightarrow a^2+2ab+b^2=c^2\)

\(\Leftrightarrow a^2+b^2-c^2=-2ab\)

Tương tự :

+) \(a^2-b^2+c^2=-2ac\)

+) \(b^2+c^2-a^2=-2bc\)

\(\Leftrightarrow A=\dfrac{a^4}{\left(-2ac\right)\left(-2ab\right)}+\dfrac{b^4}{\left(-2ab\right)\left(-2bc\right)}+\dfrac{c^4}{\left(-2bc\right)\left(-2ac\right)}\)

\(=\dfrac{a^4}{4a^2bc}+\dfrac{b^4}{4ab^2c}+\dfrac{c^4}{4abc^2}\)

\(=\dfrac{a^4bc+ab^4c+abc^4}{4a^2b^2c^2}\)

\(=\dfrac{abc\left(a^3+b^3+c^3\right)}{4a^2b^2c^2}\) (cậu tự chứng minh \(a^3+b^3+c^3=3abc\) nhé)

\(=\dfrac{3a^2b^2c^2}{4a^2b^2c^2}\)

\(=\dfrac{3}{4}\)

Vậy..

BĐT cần chứng minh tương đương với

\(\left(1-\frac{a^5-a^2}{a^5+b^2+c^2}\right)+\left(1-\frac{b^5-b^2}{b^5+c^2+a^2}\right)+\left(1-\frac{c^5-c^2}{c^5+a^2+b^2}\right)\le3\)

hay \(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Từ \(abc\ge1\) ta có:

\(\frac{1}{a^5+b^2+c^2}\le\frac{1}{\frac{a^5}{abc}+b^2+c^2}=\frac{1}{\frac{a^4}{bc}+b^2+c^2}\)

\(\le\frac{1}{\frac{2a^4}{b^2+c^2}+b^2+c^2}=\frac{b^2+c^2}{2a^4+\left(b^2+c^2\right)^2}\)

Do \(4u^2+v^2\ge4uv\Leftrightarrow4u^2+v^2\ge\frac{2}{3}\left(u+v\right)^2\)nên 

\(2a^4+\left(b^2+c^2\right)^2\ge\frac{2}{3}\left(a^2+b^2+c^2\right)^2\)

Suy ra \(\frac{1}{a^5+b^2+c^2}\le\frac{3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Tương tự ta có \(\frac{1}{b^5+c^2+a^2}\le\frac{3\left(c^2+a^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

và \(\frac{1}{c^5+a^2+b^2}\le\frac{3\left(a^2+b^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)

Cộng ba vế của các BĐT trên ta được

\(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)

Vậy \(\frac{a^5-a^2}{a^5+b^2+c^2}+\frac{b^5-b^2}{b^5+c^2+a^2}+\frac{c^5-c^2}{c^5+a^2+b^2}\ge0\)

(Dấu "="\(\Leftrightarrow a=b=c=1\))