Ta có :

<=> u³ - 3u - 2 \(\le\) v³ - 3v + 2 <=> ( u + 1 )²( u - 2 ) \(\le\) ( v - 1 )²( v + 2 )

Đặt x = u + 1 , y = v -1 thì :

BĐT <=> x³ - 3x² \(\le\) y³ + 3y² <=> x³ - y^{3 \(\le\)} 3(x² + y²)

Ta có : x - y = ( u - v ) + 2 \(\le\)2

=> ( x - y ) ( x² + xy + y² ) \(\le\)2( x² + xy + y²) = 2(x² + y²) + 2xy \(\le\) 2(x² + y²) + ( x² + y² ) = 3(x² + y² ) => x³ - y³ \(\le\) 3(x² +y² ) ( đpcm)

Dấu bằng xảy ra khi <=> x = y = 0 <=> u = -1 ; v = 1

Lời giải:

Ta thấy \(x^3+y^3+z^3\leq 9\)

\(\Leftrightarrow (x+y+z)^3-3(x+y)(y+z)(z+x)\leq 9\)

\(\Leftrightarrow 27-3[(x+y+z)(xy+yz+xz)-xyz]\leq 9\)

\(\Leftrightarrow 3(xy+yz+xz)-xyz\geq 6(\star)\)

Vì \(x,y,z\in [0;2]\Rightarrow (x-2)(y-2)(z-2)\leq 0\)

\(\Leftrightarrow xyz+4\leq 2(xy+yz+xz)\)

Mặt khác \(xyz\geq 0\rightarrow 2(xy+yz+xz)\geq 4\rightarrow xy+yz+xz\geq 2\)

Do đó \(3(xy+yz+xz)-xyz\geq 2+4+xyz-xyz=6\)

Từ đó BĐT \((\star)\) hay ta có đpcm

Dấu bằng xảy ra khi \((x,y,z)=(2,1,0)\) và các hoán vị.

a) P = sin²α  + sin²α.\(\frac{cos\text{α}}{sin\text{α}}\) + cos²α - cos²α.\(\frac{sin\text{α}}{cos\text{α}}\)

=sin²α + sinα.cosα + cos²α - cosα.sinα

=sin²α + cos²α

=1

Vậy P không phụ thuộc vào α

b) Q= -cos⁴α(2cos²α -1 -2) +sin⁴α(1 -2sin²α+2)

= -cos⁴α(cos2α -2) +sin⁴α(cos2α +2)

=-cos⁴α.cos2α +2cos⁴α +sin⁴α.cos2α +2sin⁴α

=cos2α(sin⁴α -cos⁴α) +2(sin⁴α +cos⁴α)

=cos2α [\(\left(\frac{1-cos^22\text{α}}{2}\right)^2-\left(\frac{1+cos^22\text{α}}{2}\right)^2\)]+2.[\(\left(\frac{1-cos^22\text{α}}{2}\right)^2+ \left(\frac{1+cos^22\text{α}}{2}\right)^2\)]

= -cos2α.cos2α +1+cos²2α

= -cos²2α +1+cos²2α

=1

Vậy Q không phụ thuộc vào α

ta có : \(cos^4\alpha\left(3-2cos^2\alpha\right)+sin^4\alpha\left(3-2sin^2\alpha\right)\)

\(=3cos^4\alpha-2cos^6\alpha+3sin^4\alpha-2sin^6\alpha\)

\(=3\left(sin^4\alpha+cos^4\alpha\right)-2\left(sin^6\alpha+cos^6\alpha\right)\)

\(=3\left(\left(sin^2\alpha+cos^2\alpha\right)-2sin^2\alpha.cos^2\alpha\right)-2\left(\left(sin^2\alpha+cos^2\alpha\right)^3-3sin^2\alpha.cos^2\alpha\left(sin^2+cos^2\alpha\right)\right)\)

\(=3\left(1-2sin^2\alpha.cos^2\alpha\right)-2\left(1-3sin^2\alpha.cos^2\alpha\right)\)

\(=3-6sin^2\alpha.cos^2\alpha-2+6sin^2\alpha.cos^2\alpha=1\) (không phụ thuộc vào \(\alpha\)) (đpcm)

ta có: (a-b)² \(\ge\) 0

=> a² + b² - 2ab \(\ge\) 0

=> a² +b² - ab \(\ge\) 0

=> a² +b² \(\ge\) ab

=> (a+ b)(a² +b² - ab) \(\le\) ab(a+b) (vì a\(\le0;\) b\(\le0\) nên a+b \(\le\)0)

=> a³ + b³ \(\le\) ab(a+b)

=>đpcm.

dáng lẽ phải là \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) a2 +b2 - ab ≥ ab

3: =>a^3+b^3+c^3>=3abc

=>(a+b)^3+c^3-3ab(a+b)-3abc>=0

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

=>a^2+b^2+c^2-ab-bc-ac>=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac>=0

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

Ý bạn là vầy à ?

So sánh:

\(\sqrt{2^{333}}\) và \(\sqrt{3^{222}}\)

\(2^{333}=8^{111}< 9^{111}=3^{222}\)

=>\(\sqrt{2^{333}}< \sqrt{3^{222}}\)