a) \(\frac{a+b}{2}\ge\sqrt{ab}\)

\(\Leftrightarrow\frac{a^2+2ab+b^2}{4}-ab\ge0\)

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

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

=>đpcm

Cô si

\(\frac{bc}{a}+\frac{ca}{b}\ge2\sqrt{\frac{bc}{a}\cdot\frac{ca}{b}}=2c\)

\(\frac{ca}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ca}{b}\cdot\frac{ab}{c}}=2a\)

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}\cdot\frac{bc}{a}}=2b\)

Cộng lại ta có:

\(2\left(\frac{bc}{a}+\frac{ca}{b}+\frac{ab}{c}\right)\ge2\left(a+b+c\right)\Rightarrowđpcm\)

\(12=3a+5b\ge2\sqrt{3a.5b}=2\sqrt{15ab}\Rightarrow ab\le\frac{36}{15}=\frac{12}{5}\)

Dấu " = " xảy ra khi \(3a=5b;3a+5b=12\Leftrightarrow a=2;b=\frac{6}{5}\)

Nguồn: Mr Lazy

trong câu tương tự bài của Mr Lazy đấy triều

\(12=3a+5b\ge2\sqrt{3a.5b}=2\sqrt{15ab}\Rightarrow ab\le\frac{36}{15}=\frac{12}{5}\)

Dấu "=" xảy ra khi \(3a=5b;\text{ }3a+5b=12\Leftrightarrow a=2;\text{ }b=\frac{6}{5}\)

bieu thuc:
m = a² + ab + b² - 3a - 3b + 2001

ta nhom cac hang tu:

m = (a² - 3a) + (ab) + (b² - 3b) + 2001

= (a² - 3a) + (b² - 3b) + ab + 2001

xet bieu thuc m theo a va b, ta co y tuong hoan thanh binh phuong

xet: a² - 3a = (a - 3/2)² - 9/4
xet: b² - 3b = (b - 3/2)² - 9/4

=> m = (a - 3/2)² - 9/4 + (b - 3/2)² - 9/4 + ab + 2001

= (a - 3/2)² + (b - 3/2)² + ab + 2001 - 9/2

= (a - 3/2)² + (b - 3/2)² + ab + 1996.5

de m nho nhat thi 2 binh phuong phai nho, va ab phai nho

do do ta set (a - 3/2) = x, (b - 3/2) = y => a = x + 3/2, b = y + 3/2

thay vao ta co:

m = x² + y² + (x + 3/2)(y + 3/2) + 1996.5

= x² + y² + xy + (3/2)x + (3/2)y + 9/4 + 1996.5

= x² + y² + xy + (3/2)(x + y) + 2006.75

muc tieu la tim x va y de bieu thuc nho nhat

bieu thuc chinh la: x² + y² + xy + (3/2)(x + y) + hang so

de nho nhat thi dao ham hoac thu thu

nhung ta co the thu cac gia tri x = 0, y = 0

=> a = 3/2, b = 3/2

=> m = (3/2)² + (3/2)(3/2) + (3/2)² - 3(3/2) - 3(3/2) + 2001
= 2.25 + 2.25 + 2.25 - 4.5 - 4.5 + 2001
= 6.75 - 9 + 2001 = 1998.75

=> m nho nhat la 1998.75 khi a = b = 3/2

vay:

a = 3/2, b = 3/2 thi m dat gt nho nhat la 1998.75

Áp dụng BĐT Côsi cho 2 số dương, ta có:

\(3a+5b=12\ge2\sqrt{3a.5b}=2\sqrt{15ab}\)

\(\Leftrightarrow\sqrt{15ab}\le6\)

\(\Leftrightarrow ab\le\dfrac{36}{15}\)

Dấu "=" \(\Leftrightarrow\left\{{}\begin{matrix}3a=5b\\3a+5b=12\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=\dfrac{6}{5}\end{matrix}\right.\)

cám ơn

Theo BĐT cosi ta có:

\(3a+5b\ge2\sqrt{3a\cdot5b}\)

\(\Leftrightarrow3a+5b\ge2\sqrt{15ab}\)

\(\Leftrightarrow12\ge2\sqrt{15ab}\)

\(\Leftrightarrow\sqrt{15ab}\le\dfrac{12}{2}\)

\(\Leftrightarrow\sqrt{15ab}\le6\)

\(\Leftrightarrow15ab\le36\)

\(\Leftrightarrow ab\le\dfrac{36}{15}\)

\(\Leftrightarrow ab\le\dfrac{12}{5}\)

\(\Rightarrow P\le\dfrac{12}{5}\)

Vậy: \(P_{max}=\dfrac{12}{5}\)

ôi dào !dễ ợt ! cô em mới cho học ngày hôm qua !k đi rùi em trình bày cho cách làm !