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

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

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 ~ 

b) Áp dụng bđt Holder ta có:

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

Lại có \(a^2\left(b+c\right)^2+b^2\left(c+a\right)^2+c^2\left(a+b\right)^2\le2a^2\left(b^2+c^2\right)+2b^2\left(c^2+a^2\right)+2c^2\left(a^2+b^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\ge\sqrt{\dfrac{\left(a^2+b^2+c^2\right)^3}{4\left(a^2b^2+b^2c^2+c^2a^2\right)}}\).

Ta chỉ cần chứng minh: \(\dfrac{\sqrt[4]{27\left(a^4+b^4+c^4\right)}}{2}\le\sqrt{\dfrac{\left(a^2+b^2+c^2\right)^3}{4\left(a^2b^2+b^2c^2+c^2a^2\right)}}\Leftrightarrow27\left(a^4+b^4+c^4\right)\left(a^2b^2+b^2c^2+c^2a^2\right)^2\le\left(a^2+b^2+c^2\right)^3\).

Áp dụng bđt AM - GM ta có \(27\left(a^4+b^4+c^4\right)\left(a^2b^2+b^2c^2+c^2a^2\right)^2\le\left(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\right)=\left(a^2+b^2+c^2\right)^2\).

Vậy ta có đpcm.

a) Câu này cũng tương tự: Áp dụng bđt Holder ta có:

\(\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\left(a^2+b^2+c^2\right)^3\).

Đến đây làm tương tự là ok

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..

mình mới học lớp 6 thôi

Bài 1:

\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)

\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)

Ta có:

\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)

1. Ta có $a + b + c = 0$

\(\Rightarrow\) $( a + b + c)^2 = 0$

\(\Leftrightarrow\) $a^2+b^2 +c^2 +2ab+2bc+2ac = 0
\(\Leftrightarrow\) $a^2 + b^2 + c^2 = -2(ab+bc+ac)$

Thay $a^2 + b^2 + c^2 = 2$

\(\Rightarrow\)$2 = -2(ab+bc+ac)$ \(\Rightarrow\) $ab + bc +ac = -1 $
Ta có: $(a^2+b^2+c^2) = 2$

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

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

\(\Leftrightarrow\) $a^4+b^4+c^4 + 2(a^b^2+b^2c^2+a^2c^2) = 4$ (1)
Do $2(ab+bc+ac)^2 = 2(a^2b^2+b^2c^2+a^2c^2 + 2a^2bc+2ab^2c+2abc^2)$ (2)
Từ (1)(2) => $a^4+b^4+c^4+2(ab+bc+ac)^2 - 4abc(a+b+c) = 4$(3)
Thay $(ab+bc+ac) = -1$ và $a+b+c = 0$ (4)
Từ (3)(4) => $a^4 + b^4 + c^4 +2(-1)^2 -4abc.(0) = 4 $
<=> $a^4 + b^4 + c^4 + 2 = 4 => a^4 + b^4 + c^4 = 2 $

Theo đề có \(a+b+c=0 \Rightarrow (a+b+c)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)

\(\Rightarrow ab+bc+ca=\frac{0-2}{2} = -1\) (Vì \(a^2+b^2+c^2=2\))

\(\Rightarrow (ab+bc+ca)^2=1 \)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2bc^2a+2ca^2b=1\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2 = 1\) (vì \(a+b+c=0\))

Mặt khác từ `a^2+b^2+c^2=2`

`\Rightarrow(a^2+b^2+c^2)^2=2^2`

`\Rightarrowa^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2)=4`

`\Rightarrowa^4+b^4+c^4+2.1=4`

`\Rightarrowa^4+b^4+c^4=4-2=2`