x⁴+y⁴+(x+y)⁴=x⁴+y⁴+x⁴+4x³y+6x²y²+4xy³+y⁴

=2x⁴+2y⁴+4x²y²+4x³y+4xy³+2x²y²

=2(x⁴+y⁴+2x²y²)+4xy(x²+y²)+2x²y²

=2(x²+y²)²+4xy(x²+y²)+2x²y²

=2[(x²+y²)+2xy(x²+y²)+x²y²]

=2(x²+y²+xy)² (Đpcm)

Chứng minh vế trái bằng vế phải:

\(x^4+y^4+\left(x+y\right)^4=2x^4+2y^4+4x^3y+4xy^3+6x^2y^2\)

\(=2\left(x^4+y^4+2x^3y+2xy^3+3x^2y^2\right)\)

\(=2\left(x^4+y^4+x^2y^2+2x^3y+2xy^3+2x^2y^2\right)\)

\(=2\left(x^2+y^2+xy\right)^2\)

\(\text{Chứng minh vế trái bằng vế phải: }\)

\(x^4+y^4+\left(x+y\right)^4=2x^4+2y^4+4x^3y+4xy^3+6x^2y^2\)

\(=2\left(x^4+y^4+2x^3y+2xy^3+3x^2y^2\right)\)

\(=2\left(x^4+y^4+x^2y^2+2x^3y+2xy^3+2x^2y^2\right)\)

\(=2\left(x^2+y^2+xy\right)^2\)

x⁴ + y⁴ + (x + y)⁴ = x⁴ + y⁴ + x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

= 2x⁴ + 2y⁴ + 4x²y² + 4x³y + 4xy³ + 2x²y²

= 2(x⁴ + y⁴ + 2x²y²) + 4xy(x² + y²) + 2x²y²

= 2(x² + y²)² + 4xy(x² + y²) + 2x²y²

= \(2\left [ (x^{2} + y^{2}) + 2xy(x^{2} + y^{2}) + x^{2}y^{2} \right ]\)

= 2(x² + xy + y²)² (đpcm)

x⁴ + y⁴ +(x+y)⁴ = x⁴ + y⁴ + x⁴ + 4x³y + 6x²y² +4xy³ + y⁴ = 2x⁴ +2y⁴ +4x²y²+4x³y+4xy³+2x²y²

= 2(x⁴ +y⁴ +2x²y²)+4xy(x²+y²) + 2x²y²= 2(x² + y²)² + 4xy(x² + y²) +2x²y²

=2((x² +y²) +2xy(x²+ y²) +x²y²) = 2(x² + y² + xy)² \(\Rightarrow\)  đpcm

x⁴ + y⁴ +(x+y)⁴ = x⁴ + y⁴ + x⁴ + 4x³y + 6x²y² +4xy³ + y⁴ = 2x⁴ +2y⁴ +4x²y²+4x³y+4xy³+2x²y²

= 2(x⁴ +y⁴ +2x²y²)+4xy(x²+y²) + 2x²y²= 2(x² + y²)² + 4xy(x² + y²) +2x²y²

=2((x² +y²) +2xy(x²+ y²) +x²y²) = 2(x² + y² + xy)² \(\Rightarrow\)  đpcm

a: (a+b+c)^2+a^2+b^2+c^2

=a^2+b^2+c^2+a^2+b^2+c^2+2ab+2ac+2bc

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

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

b: (x+y)^4-2(x^2+xy+y^2)^2

=(x^2+2xy+y^2)^2-2(x^2+xy+y^2)^2

=x^4+4x^2y^2+y^4+4x^3y+2x^2y^2+4xy^3-2(x^4+x^2y^2+y^4+2x^3y+2x^2y^2+2xy^3)

=-x^4-y^4

=>ĐPCM

Hằng đẳng thức ???

Áp dụng BĐT \(x^2+y^2\ge2xy\) ta có:

\(\frac{x^4+y^4}{2}\ge\frac{\left(x^2\right)^2+\left(y^2\right)^2}{2}\ge\frac{2x^2y^2}{2}=x^2y^2\)

Tương tự cho 2 BĐT còn lại cũng có;

\(\frac{y^4+z^4}{2}\ge y^2z^2;\frac{z^4+x^4}{2}\ge x^2z^2\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=\frac{x^4+y^4}{2}+\frac{y^4+z^4}{2}+\frac{z^4+x^4}{2}\ge x^2y^2+y^2z^2+z^2x^2=VP\)

Khi \(x=y=z\)

Áp dụng bđt Cô si cho 2 số không âm, ta có:

\(\hept{\begin{cases}\frac{x^4+y^4}{2}\ge\sqrt{x^4y^4}=x^2y^2\\\frac{y^4+z^4}{2}\ge\sqrt{y^4z^4}=y^2z^2\\\frac{z^4+x^4}{2}\ge\sqrt{z^4x^4}=z^2x^2\end{cases}}\)

\(\Rightarrow\frac{x^4+y^4}{2}+\frac{y^4+z^4}{2}+\frac{z^4+x^4}{2}\ge x^2y^2+y^2z^2+z^2x^2\)

quốc huy khìn hử, đề có cho số dương hay ko âm đâu mà Cô si

Ta có: \(x^4+y^4+\left(x+y\right)^4\)\(=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(=2x^4+2y^4+4x^2y^2+4x^3y+4xy^3+2x^2y^2\)

\(=2\left(x^4+y^4+2x^2y^2\right)+4xy\left(x^2+y^2\right)+2x^2y^2\)

\(=2\left(x^2+y^2\right)^2+4xy\left(x^2+y^2\right)+2x^2y^2\)

\(=2\left[\left(x^2+y^2\right)+2xy\left(x^2+y^2\right)+x^2y^2\right]\)

\(=2\left(x^2+xy+y^2\right)^2\left(dpcm\right)\)