Question

a) Chứng minh rằng: (ax+by)²\(\le\)(a²+b²)(x²+y²)

b) Cho x²+y²=52. Tìm GTLN của B = 2x+3y

Answer 1

a)  \(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Leftrightarrow\)\(\left(ax\right)^2+2axby+\left(by\right)^2\le\left(ax\right)^2+\left(ay\right)^2+\left(bx\right)^2+\left(by\right)^2\)

\(\Leftrightarrow\)\(2axby\le\left(ay\right)^2+\left(bx\right)^2\)

\(\Leftrightarrow\)\(\left(ay\right)^2-2axby+\left(bx\right)^2\ge0\)

\(\Leftrightarrow\)\(\left(ay-bx\right)^2\ge0\)

Dấu "=" xảy ra   \(\Leftrightarrow\)  \(\frac{a}{x}=\frac{b}{y}\)

Answer 2

Áp dụng BĐT Cauchy–Schwarz ta có:

\(B^2=\left(2x+3y\right)^2\le\left(2^2+3^2\right)\left(x^2+y^2\right)=676\)

=>  \(B\le26\)

Vậy  MAX   \(B=26\)  khi    \(\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\2x+3y=26\end{cases}}\)<=>  \(\hept{\begin{cases}x=4\\y=6\end{cases}}\)