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áp dụng BĐT (a - b)² ≥ 0 → a² + b² ≥ 2ab ta có:
x² + y² ≥ 2xy
x² + 1 ≥ 2x
y² + z² ≥ 2yz
y² + 1 ≥ 2y
z² + x² ≥ 2xz
z² + 1 ≥ 2z
Cộng theo vế → 3(x² + y² + z²) + 3 ≥ 2(x + y + z + xy + yz + zx) = 2.6 = 12
→ x² + y² + z² ≥ 9/3 = 3
→ đpcm (dấu = xảy ra khi x = y = z = 1)
Bài 1:
Ta chứng minh bất đẳng thức phụ sau:
\(a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2bc+2ca\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(luôn đúng).
Áp dụng vào bài toán:
\(x^2+y^2+z^2\ge xy+yz+zx\)\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)(1)
Sử dụng BĐT Cauchy, ta được:
\(x^2+1\ge2x;\)\(y^2+1\ge2y;\)\(z^2+1\ge2z\)
Cộng theo vế: \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)(2)
Cộng (1) với (2) theo vế: \(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)
Thay \(x+y+z+xy+yz+zx=6\)
Suy ra: \(3\left(x^2+y^2+z^2\right)+3\ge12\Leftrightarrow x^2+y^2+z^2\ge3\)(đpcm).
Bài 2:
Ta có: \(a^4+b^4-a^3b-ab^3=a^3\left(a-b\right)+b^3\left(b-a\right)\)
\(=a^3\left(a-b\right)-b^3\left(a-b\right)=\left(a-b\right).\left(a^3-b^3\right)\)
\(=\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(luôn đúng)
Suy ra \(a^4+b^4\ge a^3b+ab^3\)(1)
Áp dụng BĐT Cauchy, ta có: \(a^4+b^4\ge2\sqrt{a^4b^4}=2a^2b^2\)(2)
Cộng (1) với (2) theo vế, ta được:
\(2\left(a^4+b^4\right)\ge ab^3+a^3b+2a^2b^2\)(đpcm).
1)2( x2+y2+z2)>=2(xy+yz+xz )
x2+1>=2x
y2+1>=2y
z2+1>=2z
=>3(x2+y2+z2)>= 2(x+y+z+xy+yz+xz)=12
=> x2+y2+z2>=3
2) ta co a^4+b^4 >=2a^b^2 voi moi a,b
lai co a^4 +b^4 - ab^3-a^3b
=a^3(a-b)-b^3(a-b)
=(a-b)(a^3-b^3)
=(a-b)^2(a^2+b^2+ab)>=0 voi moi a,b
=> 2(a^4+b^4)>= ab^3+a^3b+2a^2b^2 voi moi a,b
\(1.\)
\(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^3z-x^2z^2+x^2yz-xyz^2\)
\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)
\(=\left(x^2z-xyz\right)\left(x-z\right)\)
\(=xz\left(x-y\right)\left(x-z\right)\)
\(2.\)
\(x^2-\left(a+b\right)xy+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x^2-bxy-axy+aby^2\)
\(=x\left(x-by\right)-ay\left(x-by\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
\(3.\)
\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)
\(=abx^2+aby^2+a^2xy+b^2xy\)
\(=abx^2+b^2xy+a^2xy+aby^2\)
\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)
\(=\left(ax+by\right)\left(bx+ay\right)\)
\(4.\)
\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)
\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)
\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)
\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)
\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)
\(5.\)
\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)
\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)
\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)
\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)
\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)
\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(6.\)
\(16x^2-40xy+2y^2\)
\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)
\(=\left(4x-5y\right)^2\)
\(7.\)
\(25x^4-10x^2y+y^2\)
\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)
\(=\left(5x^2+y\right)^2\)
\(8.\)
\(-16x^4y^6-24x^5y^5-9x^6y^4\)
\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)
\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)
\(=\left(4x^2y^3+3x^3y^2\right)^2\)
\(9.\)
\(16x^2-4y^2-8x+1\)
\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)
\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)
\(=\left(4x+1\right)^2-\left(2y\right)^2\)
\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)
\(10.\)
\(49x^2-25+42xy+9y^2\)
\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)
\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)
\(=\left(7x+3y\right)^2-5^2\)
\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)
Bài 1 :
Theo BĐT cô - si ta có :
\(x^2+y^2\ge2xy\)
\(x^2+1\ge2x\)
\(y^2+z^2\ge2yz\)
\(y^2+1\ge2y\)
\(z^2+x^2\ge2zx\)
\(z^2+1\ge2z\)
Cộng vế theo vế ta được :
\(3\left(x^2+y^2+z^2\right)+3\ge2\left(x+y+z+xy+yz+zx\right)\)
\(\Leftrightarrow3\left(x^2+y^2+z^2\right)+3\ge12\)
\(\Leftrightarrow x^2+y^2+z^2\ge3\left(đpcm\right)\)
Bài 2 :
Ta có :
\(2\left(a^4+b^4\right)\ge ab^3+a^3b+2a^2b^2\)
\(\Leftrightarrow a^4+b^4+a^4+b^4-ab^3-a^3b-2a^2b^2\ge0\)
\(\Leftrightarrow\left(a^4-a^3b-ab^3+b^4\right)+\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)+\left(a^2-b^2\right)^2\ge0\)
2)\(2\left(a^4+b^4\right)\ge ab^3+a^3b+2a^2b^2\)
\(\Leftrightarrow\left(a^4-2a^2b^2+b^4\right)+a^4+b^4-ab^3-a^3b\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+a^3\left(a-b\right)+b^3\left(b-a\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a-b\right)\left(a^3-b^3\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a-b\right)\left(a-b\right)\left(a^2+ab+b^2\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(luôn đúng)
Bài 1: 4a2-4ab+b2-9a2b2
=(2a)2-2.2a.b+b2-(3ab)2
=(2a-b)2-(3ab)2
=(2a-b-3ab)(2a-b+3ab)
a/ (4a2-4ab+b2)-9a2b2
= (2a-b)2-(3ab)2
= (2a-b-3ab) (2a-b+3ab)
a) đề <=> 2(x^4 +y^4+z^4)-2(x^2y^2 +y^2z^2+z^2x^2)>=0
<=>(x^2-y^2)^2 +(x^2-z^2)^2 + (y^2-z^2)^2>= 0 (đúng)
=> bất đẳng thức ở đề đúng
b) Cmtt câu a rồi suy ra (a-b)^2 + (b-c)^2 + (c-a)^2>=0