Ta có: \(a+b+c=0\)

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

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)

Mặt khác: \(a^2\ge0\forall a;b^2\ge0\forall b;c^2\ge0\forall c\)

\(\Rightarrow a^2+b^2+c^2\ge0\) 

Suy ra: \(2ab+2bc+2ac=0\)

\(\Rightarrow2\left(ab+bc+ac\right)=0\)

\(\Rightarrow ab+bc+ac=0\Leftrightarrow2\left(ab+bc+ac\right)^2=0\) (1)

Lại có: \(a^4+b^4+c^4\)

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

\(=0-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2+2\left(ab+bc+ac\right)-2\left(ab+bc+ac\right)\right]\)

\(=-2\left(ab+bc+ac\right)^2-4\left(ab+bc+ac\right)\)

\(=0\) (2)

Từ (1) và (2) \(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ac\right)^2=0\)

hay \(a^4+b^4+c^4=2\left(ab+ac+bc\right)^2\)

Kiểm tra hộ mình xem có đúng không ạ!

\(A=3+3^2+3^3+...+3^{100}\)

\(\Leftrightarrow3A=3^2+3^3+3^4+3^5+....+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+3^4+3^5+...+3^{101}\right)-\left(3+3^2+3^3+3^4+...+3^{100}\right)\)

\(\Leftrightarrow2A=3^{101}-3\)

\(\Leftrightarrow A=\frac{3^{101}-3}{2}< 3^{100}-1\)

\(\Leftrightarrow A< B\)

a. tính A = 3+3^2+3^3+3^4+.....+3^100

3A=3^2+3^3+3^4+3^5+....+3^100

3A-A=(3^2+3^3+3^4+....+3^101)-(3+3^2+3^3+3^4+.....+3^100)=3^101-3=3^100

mà B=3^100-1 => A<B

\(A=1+4+4^2+...+4^{99}\)

\(\Leftrightarrow4A=4+4^2+4^3+...+4^{100}\)

\(\Leftrightarrow3A=4^{100}-1\)

\(\Leftrightarrow A=\frac{4^{100}-1}{3}< \frac{4^{100}}{3}\)

hay A<B (đpcm)

Vì (a + 3)(b - 4) - (a - 3)(b + 4) = 0

<=> (a+3)(b - 4) = (a-3)(b + 4)

<=> \(\frac{a+3}{b+4}=\frac{a-3}{b-4}\)(t/c tỉ lệ thức)

=> \(\frac{a+3}{b+4}=\frac{a-3}{b-4}=\frac{a+3+a-3}{b+4+b-4}=\frac{a+3-a+3}{b+4-b+4}\)

=> \(\frac{2a}{2b}=\frac{6}{8}\)

=> \(\frac{a}{b}=\frac{3}{4}\)

=> \(\frac{a}{3}=\frac{b}{4}\)

Áp dụng Côsi:

\(a^4+a^4+a^4+1\ge4\sqrt[4]{\left(a^4\right)^3}=4a^3\)

\(\Rightarrow3\left(a^4+b^4+c^4+d^4\right)\ge4\left(a^3+b^3+c^3+d^3\right)-1\)

Ta chứng minh: \(a^3+b^3+c^3+d^3\ge4\)

Theo Côsi: \(a^3+1+1\ge3\sqrt[3]{a^3}=3a\)

\(\Rightarrow a^3+b^3+c^3+d^3+2.4\ge3\left(a+b+c+d\right)=3.4\)

\(\Rightarrow a^3+b^3+c^3+d^3\ge4\)

\(\Rightarrow3\left(a^4+b^4+c^4+d^4\right)\ge4\left(a^3+b^3+c^3+d^3\right)-4\ge3\left(a^3+b^3+c^3+d^3\right)\)

\(\Rightarrow a^4+b^4+c^4+d^4\ge a^3+b^3+c^3+d^3\)

Ta có

\(\left(a+b\right)^2=a^2+b^2+2ab=1\Rightarrow a^2+b^2=1-2ab\) (1)

Ta có

\(\left(a+b\right)^4=\left(a^2+b^2+2ab\right)^2=\)

\(=a^4+b^4+4a^2b^2+2a^2b^2+4ab^3+4a^3b=\)

\(=a^4+b^4+6a^2b^2+4ab\left(a^2+b^2\right)=1\)

\(\Rightarrow a^4+b^4=1-6a^2b^2-4ab\left(1-2ab\right)=\)

\(=1-6a^2b^2-4ab+8a^2b^2=\)

\(=1+2a^2b^2-4ab\) (2)

Ta có

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

\(=1-2ab-ab=1-3ab=1\Rightarrow ab=0\)

Thay \(ab=0\) vào (1) và (2)

\(a^2+b^2=1-2ab=1\)

\(a^4+b^4=1+2a^2b^2-4ab=1\)

\(\Rightarrow a^2+b^2=a^4+b^4\)

Ta chứng minh: \(\dfrac{a^4+b^4}{a^3+b^3}\ge\dfrac{a+b}{2}\Leftrightarrow2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\Leftrightarrow2a^4+2b^4\ge a^4+ab^3+b^4+ba^3\)

\(\Leftrightarrow a^4+b^4\ge ab^3+ba^3\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\Leftrightarrow\left(a-b\right)\left(a^3-b^3\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)Bất đẳng thức cuối luôn đúng nên ta có điều phải chứng minh. Áp dụng vào bài, ta có:

\(\dfrac{a^4+b^4}{a^3+b^3}+\dfrac{b^4+c^4}{b^3+c^3}+\dfrac{c^4+a^4}{c^3+a^3}\ge\dfrac{a+b}{2}+\dfrac{b+c}{2}+\dfrac{c+a}{2}=2018\)