Question

Tìm giá trị nhỏ nhất của:

A= x³+y³+xy biết x+y =1

Answer 1

Cách khác :

A = x³ + y³ + xy

A = ( x + y)( x² - xy + y²) + xy

A = x² - xy + y² + xy

A = x² + y²

Ta có : x + y = 1

⇔ ( x + y)² = 1

⇔ x² + 2xy + y² = 1 ( 1)

Mà : ( x - y)² ≥ 0 ∀xy

⇔ x² - 2xy + y² ≥ 0 ( 2)

Cộng ( 1 ; 2) Ta được : 2( x² + y²) ≥ 1

⇔ x² + y² ≥ \(\dfrac{1}{2}\)

⇒ A_MIN = \(\dfrac{1}{2}\) ⇔ x = y = \(\dfrac{1}{2}\)

Answer 2

ta có:

x^3 + y^3 + xy = x^3 + y^3 + 3xy (x + y) + xy - 3xy (x+y = 1)

= (x + y)^3 - 2xy = 1 - 2xy

áp dụng bđt cauchy:

2xy ≤ \(\dfrac{\left(x+y\right)^2}{2}=\dfrac{1}{2}\)

<=> -2xy ≥ -1/2

<=>> 1 -2xy ≥ 1 - 1/2 = 1/2

dấu "=" xảy ra khi x = y = 1/2

gtnn = 1/2 khi x = y = 1/2

Answer 3

A=x^3+y^3+xy

=(x+y)[(x+y)^2-3xy]+xy

=1-2xy

x+y=1=>-y=x-1

A=1+2x(x-1)

A=2x^2-2x+1

A=(√2x-1/√2)^2+1-1/2≥1/2

khi x=y=1/2

Answer 4

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

\(\text{Lại có: }x+y=1\\ \Rightarrow y=1-x\)

\(\Rightarrow A=x^2+\left(1-x\right)^2\\ =x^2+1-2x+x^2\\ =2x^2-2x+2\\ =2x^2-2x+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(2x^2-2x+\dfrac{1}{2}\right)+\dfrac{1}{2}\\ =2\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{1}{2}\\ =2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\)

Do \(2\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

\(\Rightarrow2\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\forall x\)

Dấu "=" xảy ra khi:

\(2\left(x-\dfrac{1}{2}\right)^2=0\\ \Leftrightarrow x-\dfrac{1}{2}=0\\ \Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(A_{Min}=\dfrac{1}{2}\) khi \(x=\dfrac{1}{2}\)