Nếu cái j?

đặt x/a=y/b=z/c=k

=>x=a.k,

y=b.k

z=c.k

=>(a^2k^2+b^2k^2+c^2k^2)(a^2+b^2+c^2)=k^2.(a^2+b^2+c^2)^2(1)

(ax+by+cz)^2=(a.a.k+b.b.k+c.c.k)^2=(a^2.k+b^2.k+c^2.k)^2

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

từ (1)(2)=> nếu x/a=y/b=z/c thì (x² + y² + z²) (a² + b² + c²) = (ax + by + cz)²

=>

\(1.a,\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=\left(ac\right)^2+2abcd+\left(bd\right)^2+\left(ad\right)^2-2abcd+\left(bc\right)^2\)

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

\(b,\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ad-bc\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow-\left(ad-bc\right)^2\le0\left(luôn-đúng\right)\)

\(dấu"='\) \(xảy\) \(ra\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

\(c2:x+y=2\Rightarrow\left(x+y\right)^2=4\)

\(\Rightarrow\left(x+y\right)^2+\left(x-y\right)^2\ge4\)

\(\Leftrightarrow x^2+2xy+y^2+x^2-2xy+y^2\ge4\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge4\Leftrightarrow x^2+y^2\ge2\)

\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=y=1\)

Câu 1:

a)Ta có (ac+bd)²+(ad-bc)²=(ac)²+2abcd+(bd)²+(ad)²-2abcd+(bc)²

                                          =(ac)²+(bd)²+(ad)²+(bc)²

                                          =a²(c²+d²)+b²(c²+d²)

                                          =(a²+b²)(c²+d²) (đpcm)

b)Ta có (ac+bd)² = (ac)²+2abcd+(bd)²

Lại có (a²+b²)(c²+d²) = (ac)²+(bd)²+(ad)²+(bc)²

Ta có (ac+bd)² ≤  (a²+b²)(c²+d²) 

<=>(a²+b²)(c²+d²) - (ac+bd)² ≥ 0

<=>(ac)²+(bd)²+(ad)²+(bc)²-[(ac)²+2abcd+(bd)²]

<=>(ad)² - 2abcd +(bc)² ≥ 0

<=>(ad-bc)² ≥ 0 (Luôn đúng) => đpcm

Câu 2:

Áp dụng BĐT Bunhiacôpxki, ta có (x+ y)² ≤ (x² + y²)(1² + 1²) => 4  2.S => 2  S

Dấu ''='' xảy ra <=> x=y=1

Vậy Min S=2 <=> x=y=1

cứu

1)chứng minh cái j ???

2)\(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(=a^2c^2+b^2d^2+2abcd+a^2d^2-2abcd+b^2c^2\)

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

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

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

b)Ta có: 

\(\left(ab+cd\right)^2\le\left(a^2+c^2\right)\left(b^2+d^2\right)\)

\(\Leftrightarrow a^2b^2+c^2d^2+2abcd\le a^2b^2+a^2d^2+b^2c^2+c^2d^2\)

\(\Leftrightarrow a^2d^2+b^2c^2-2abcd\ge0\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\)(Đpcm)

c)Áp dụng Bđt Bunhiacopxki ta có:

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x+y\right)^2=2^2=4\)

\(\Rightarrow2\left(x^2+y^2\right)\ge4\)

\(\Rightarrow x^2+y^2\ge2\)\(\Rightarrow S\ge2\)

Dấu = khi \(x=y=1\)

a: \(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

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

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

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

b: Bạn ghi lại đề đi bạn

a)Ta có:

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

Do \(\left(a-b\right)^2\ge0\),nên\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

b)Xét \(\left(a+b+c\right)^2+\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\)

Khai triển và rút gọn ta được:\(3\left(a^2+b^2+c^2\right)\)

Vậy \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

10. a) Ta có : (a + b)2 + (a – b)2 = 2(a2 + b2). Do (a – b)\(^2\) ≥ 0, nên (a + b)\(^2\) ≤ 2(a2 + b2).

b) Xét : (a + b + c)\(^2\) + (a – b)\(^2\) + (a – c)\(^2\) + (b – c)\(^2\)

. Khai triển và rút gọn, ta được : 3(a\(^2\) + b\(^2\) + c\(^2\)).

Vậy : (a + b + c)\(^2\) ≤  3( a\(^2\) + b\(^2\) + c\(^2\)).

Cách khác : Biến đổi tương đương

a, \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\)

\(\Leftrightarrow a^2-2ab+b^2\ge0\)

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

b, \(\left(a+b+c\right)^2\le3\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\le3a^2+3b^2+3c^2\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

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