Ta có: 4x² + 12xy + 10y² + 4x + 4y + 2 = 0

<=> (4x² + 12xy + 9y²) + 2(2x + 3y) + 1 + (y² - 2y + 1) = 0

<=> (2x + 3y)² + 2(2x + 3y) + 1 + (y - 1)² = 0

<=> (2x + 3y + 1)² + (y - 1)² = 0

<=> \(\hept{\begin{cases}2x+3y+1=0\\y-1=0\end{cases}}\)

<=> \(\hept{\begin{cases}x=-\frac{1+3y}{2}\\y=1\end{cases}}\)

<=> \(\hept{\begin{cases}x=-2\\y=1\end{cases}}\)(tm)

Khi đó: P = \(\frac{x^2+y^2+xy}{3xy}=\frac{\left(-2\right)^2+1^2-2.1}{3.\left(-2\right).1}=-\frac{1}{2}\)

CÓ AI KẾT BẠN VS TUI KO???

ko ai kết bạn vs tui à

Đặt a² = x; b² = y; c² = z

Khi đó, ta có: (x + y)(y + z)(z + x) \(\ge\)xyz

<=> (xy + xz + y² + yz)(z + x) - 8xyz \(\ge\)0

<=> xyz + xz² + y²z + yz² + x²y + x²z + y²x + xyz - 8xyz \(\ge\)0

<=> (xz² +xy²) + (y²z + zx²) + (yz² + yx²) - 6xyz \(\ge\)0

<=> (xz² - 2xyz + xy²) + (y²z + zx² - 2xyz) + (yz² + yx² - 2xyz) \(\ge\)0

<=> x(z² - 2yz + y²) + z(y² + x² - 2xy) + y(z² + x² - 2xz) \(\ge\) 0

<=> x(z - y)² + z(y - x)² + y(z - x)² \(\ge\)0

hay a²(c² - b²)² + c²(b² - a²)² + b²(c² - a²)² \(\ge\)0 (luôn đúng với mọi a;b;c)

=> Đpcm

Đặt \(a^2;b^2;c^2\rightarrow x;y;z\left(x;y;z\ge0\right)\)

Khi đó bài toán trở thành \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)

\(< =>\left(x+y\right)\left(y+z\right)\left(z+x\right)-8xyz\ge0\)

\(< =>a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)*đúng*

Dấu "=" xảy ra khi và chỉ khi \(x=y=z\)hay \(a^2=b^2=c^2\)

Ta có: \(A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

\(\ge\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{\left(x+y\right)^2}{2}}=\frac{4}{\left(x+y\right)^2}+\frac{2}{\left(x+y\right)^2}\)

\(=\frac{6}{\left(x+y\right)^2}=6\)

Đẳng thức xảy ra khi \(x=y=\frac{1}{2}\)

Bài làm:

Ta có: \(x+y\ge2\sqrt{xy}\)(bất đẳng thức Cauchy)

\(\Leftrightarrow\sqrt{xy}\le\frac{x+y}{2}\)

\(\Leftrightarrow xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\)

Áp dụng bất đẳng thức Cauchy Schwars ta được:

\(A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\)

\(\ge\frac{\left(1+1\right)^2}{x^2+2xy+y^2}+\frac{1}{2.\frac{1}{4}}=\frac{4}{\left(x+y\right)^2}+\frac{1}{\frac{1}{2}}\)

\(=\frac{4}{1^2}+2=6\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

1. Áp dụng BĐT Cauchy dạng Engle, ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(a+b+c\right)\left(\frac{9}{a+b+c}\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

\(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)

\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)

Vì a, b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\left(\frac{a+b}{3}-1\right)\le0\Leftrightarrow a+b\le3\)

\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+2b+\frac{8}{a}+\frac{2}{b}-\left(a+b\right)\ge8+4-3=9\)

Áp dụng BĐT Cauchy cho a ; b dương

Dấu "=" xảy ra \(\Leftrightarrow a=2;b=1\)

để sai phải không ạ ? tìm Max chứ 

VT = a³ + b³ + c³ - 3abc = (a + b)(a² - ab + b²) + c³ - 3abc

= (a + b)(a² + 2ab + b² - 3ab) + c³ - 3abc

= (a + b)³ - 3ab(a + b) + c³ - 3abc

= (a + b+ c)[(a + b)² - c(a + b) + c²] - 3ab(a + b+  c)

= (a + b + c))(a² + 2ab + b² - ac - bc + c² - 3abc)

= (a + b + c)(a² + b² + c² - ab - ac - bc) = VP

=> ĐPCM

Sửa đề :

VP= (a+b+c)(a²+b²+c²-ab-bc-ca)

     =a³+ab²+ac²-a²b-abc-ca²+ba²+b³+bc²-ab²-b²c-abc+ca²+cb²+c³-abc-bc²-c²a

     =a³+b³+c³-3abc

Cách này đỡ phức tạp hơn cách của edogawa conan

Bài làm:

Ta có: \(x^2+4y^2+z^2-2x-6z+8y+15\)

\(=\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)+1\)

\(=\left(x-1\right)^2+4\left(y+1\right)^2+\left(z-3\right)^2+1\ge1>0\left(\forall x,y,z\right)\)

x² + 4y² + z² - 2x - 6z + 8y + 15 

= ( x² - 2x + 1 ) + ( 4y² + 8y + 4 ) + ( z² - 6z + 9 ) + 1

= ( x - 1 )² + ( 2y + 2 )² + ( z - 3 )² + 1 ≥ 1 > 0 ∀ x,y,z ( đpcm )

Bài làm:

a) \(x^2-6x+4=\left(x^2-6x+9\right)-5=\left(x-3\right)^2-\left(\sqrt{5}\right)^2\)

\(=\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\)

b) \(x^2-4x+3=x^2-x-3x+3=\left(x-1\right)\left(x-3\right)\)

c) \(6x^2-5x+1=6x^2-3x-2x+1=\left(2x-1\right)\left(3x-1\right)\)

d) \(3x^2+13x-10=3x^2+15x-2x-10=\left(x-5\right)\left(3x-2\right)\)