Question

Cho a(a-1) + b(b-1) + c(c-1)<= 4/3

Tìm max A=a+b+c

Answer 1

\(\left(2a-1\right)^2+\left(2b-1\right)^2+\left(2c-1\right)^2\le\dfrac{25}{3}\)(1)

\(x^2+y^2+z^2\le\dfrac{25}{3}\)(2)

\(\dfrac{\left(x+y+z\right)^2}{3}\le\left(x^2+y^2+z^2\right)\)(3) {cơ bản)

Từ (2) và (3)\(\Rightarrow x+y+z\le\sqrt{25}=5\) (4)

Từ (1) và (4) \(\Rightarrow\left(2a-1\right)+\left(2b-1\right)+\left(2c-1\right)\le5\)

\(\Rightarrow\left(a+b+c\right)\le\dfrac{\left(5+3\right)}{2}=4\\ \)

A_max=4

Đẳng thức khi:

\(\left(1\right)\&\left(3\right)\Rightarrow\left[\begin{matrix}a=b=c=\dfrac{4}{3}\\a=b=c=-\dfrac{1}{3}\end{matrix}\right.\)

Answer 2

Hint: Maximize=\(4\) at \(a=b=c=\dfrac{1}{3}\)

P/s:I think it's very easy with you and i think it will be better if you test methods