\(a^2+4b^2+4c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow\left(a-2b+2c\right)^2\ge0\forall a,b,c\)

\("="\Leftrightarrow b=\dfrac{a}{2}+c\)

Bài 1 : 

a) \(x^2+y^2\)

\(\Leftrightarrow x^2+2xy+y^2-2xy\)

\(\Leftrightarrow\left(x+y\right)^2-2xy=\left(-3\right)^2-2.\left(-28\right)=65\)

b) \(x^3+y^3\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(\Leftrightarrow\left(x+y\right)\left(x^2+2xy+y^2-3xy\right)\)

\(\Leftrightarrow\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]=\left(-3\right)\left[\left(-3\right)^2-3.\left(-28\right)\right]=-279\)

c) \(x^4+y^4\)

\(\Leftrightarrow\left(x+y\right)^4-4x^3y-4xy^3-6x^2y^2=\left(-3\right)^4-4\left(-28\right).65-6\left(-28\right)^2=2657\)

Bài 3:

Có:    \(x^3+y^3+z^3=\left(x+y\right)^3-3xy\left(x+y\right)+z^3\)

=>     \(x^3+y^3+z^3=\left(-z\right)^3-3xy.-z+z^3\)

=>     \(x^3+y^3+z^3=-z^3+z^3+3xyz=3xyz\)

=> TA CÓ ĐPCM.

VẬY      \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Bài 2 :

a) Giả sử  \(a^2+b^2+c^2+d^2\ge ab+ac+ad\)

\(\Leftrightarrow a^2+b^2+c^2+d^2-ab-ac-ad\ge0\)

\(\Leftrightarrow4a^2+4b^2+4c^2+4d^2-4ab-4ac-4ad\ge0\)

\(\Leftrightarrow\left(a-2b\right)^2+\left(a-2c\right)^2+\left(a-2d^2\right)\ge0\)( luôn đúng )

\(\RightarrowĐPCM\)

b) Sửa đề : \(a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Ta có : \(\left(a+2c\right)^2\ge0\Leftrightarrow a^2+4c^2\ge-4ac\left(1\right)\)

Áp dụng BĐT Cô - si ta có :

\(\hept{\begin{cases}a^2+4b^2\ge4ab\left(2\right)\\4b^2+4c^2\ge8bc\left(3\right)\end{cases}}\)

(1) + (2) + (3) 

\(\Leftrightarrow2a^2+8b^2+8c^2\ge4ab-4ac+8bc\)

\(\Leftrightarrow2\left(a^2+4b^2+4c^2\right)\ge4\left(ab-ac+2bc\right)\)

\(\Leftrightarrow a^2+4b^2+4c^2\ge2ab-2ac+4bc\)

Nhầm , sorry bạn nha , mk làm lại nè

a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

⇔ a² - 4ab + 4b² + 4ac - 8bc + 4c² ≥ 0

⇔ ( a - 2b)² + 4c( a - 2b) + 4c² ≥ 0

⇔ ( a - 2b + 2c)² ≥ 0 ( luôn đúng ∀abc)

\(a^2+4b^2+4c^2\ge4ab-4ac+8bc\\ \Leftrightarrow a^2+4b^2+4c^2-4ab+4ac-8bc\ge0\\ \Leftrightarrow\left(a-2b+2c\right)^2\ge0\)

Luôn đúng với \(\forall x\in R\)

a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

⇔ a² - 4ab + 4b² - 4ac + 8bc + 4c² ≥ 0

⇔ ( a - 2b)² - 4c( a - 2b) + 4c² ≥ 0

⇔ ( a - 2b - 2c)² ≥ 0 ( luôn đúng ∀a,b,c )

a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

⇔ a² + 4b² + 4c² - 4ab + 4ac - 8bc ≥ 0

⇔ (a - 2b + 2c)² ≥ 0 (đúng ∀abc)

Vậy a² + 4b² + 4c² ≥ 4ab - 4ac + 8bc

Xét hiệu (a^2 + 4b^2 + 4c^2)-( 4ab-4ac+8bc )

= (a^2-4ab+4b^2) + 4c^2 + (4ac-8bc)

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

=(a-2b+2c)^2 >=0

Vậy a^2 + 4b^2 + 4c^2 >=  4ab-4ac+8bc

hok tốt

Tất cả các câu này đều có thể chứng minh bằng phép biến đổi tương đương:

a.

\(\Leftrightarrow a^{10}+b^{10}+a^4b^6+a^6b^4\le2a^{10}+2b^{10}\)

\(\Leftrightarrow a^{10}-a^6b^4+b^{10}-a^4b^6\ge0\)

\(\Leftrightarrow a^6\left(a^4-b^4\right)-b^6\left(a^4-b^4\right)\ge0\)

\(\Leftrightarrow\left(a^6-b^6\right)\left(a^4-b^4\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)\left(a^4+a^2b^2+b^4\right)\left(a^2-b^2\right)\left(a^2+b^2\right)\ge0\)

\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^2+b^2\right)\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

b.

\(\Leftrightarrow\left(\dfrac{a^2}{4}+b^2+c^2-ab+ac-2bc\right)+b^2-2b+1+c^2\ge0\)

\(\Leftrightarrow\left(\dfrac{a}{2}-b+c\right)^2+\left(b-1\right)^2+c^2\ge0\) (luôn đúng)

c.

\(\Leftrightarrow a^2+4b^2+4c^2-4ab-8bc+4ac\ge0\)

\(\Leftrightarrow\left(a-2b+2c\right)^2\ge0\) (luôn đúng)

d.

\(\Leftrightarrow4a^4-8a^3+4a^2+a^2-2a+1\ge0\)

\(\Leftrightarrow\left(2a^2-2a\right)^2+\left(a-1\right)^2\ge0\) (luôn đúng)

cac giup minh di minh sap phai nop roi

a²+4b²⁺4c²>= 4ab-4ac+8bc

a²+4b²⁺4c² - 4ab +4ac-8bc

(a² - 4ab+4b²)+4c²+(4ac-8bc>=0)

suy ra (a-2b²)+2.2c.(a-2b)+(2c)²

(a-2b+2c)²>=0

dau = xảy ra khi va chỉ khi a+2c=2b

a²+4b²+4c²>= 4ab-4ac+8bc(dpcm)

ban giai day du cho minh di. minh lam de nop ma

a/ \(\frac{4bc-a^2}{bc+2a^2}.\frac{4ab-c^2}{ab+2c^2}.\frac{4ac-b^2}{ac+2b^2}\)

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

\(=\frac{-\left(b-c\right)^2}{\left(c+2b\right)\left(b+2c\right)}.\frac{-\left(c+2b\right)^2}{-\left(b-c\right)\left(b+2c\right)}.\frac{-\left(b+2c\right)^2}{\left(b-c\right)\left(c+2b\right)}=1\)