\(A=x^2+y^2\) hả bạn?

Áp dụng Bunyakovsky, ta có :

\(\left(1+1\right)\left(x^2+y^2\right)\ge\left(x.1+y.1\right)^2=1\)

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

=> \(Min_C=\frac{1}{2}\Leftrightarrow x=y=\frac{1}{2}\)

Mấy cái kia tương tự 

mk copy trên trang này

https://lazi.vn/edu/exercise/311935/cho-cac-so-thoa-man-2x-3y-13-tim-gia-tri-nho-nhat-cua-q

\(2x+3y=13\Rightarrow y=\dfrac{13-2x}{3}\)

\(Q=x^2+\left(\dfrac{13-2x}{3}\right)^2=\dfrac{13}{9}x^2-\dfrac{52}{9}x+\dfrac{169}{9}\)

\(Q=\dfrac{13}{9}\left(x-2\right)^2+13\ge13\)

\(Q_{min}=13\) khi \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{49}{16}\)

\(M_{min}=\dfrac{49}{16}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{\sqrt{7}};\dfrac{2}{\sqrt{14}};\dfrac{2}{\sqrt{7}}\right)\)

cm bn

\(M=\dfrac{\dfrac{1}{16}}{x^2}+\dfrac{\dfrac{1}{4}}{y^2}+\dfrac{1}{z^2}\ge\dfrac{\left(\dfrac{1}{4}+\dfrac{1}{2}+1\right)^2}{x^2+y^2+z^2}=\dfrac{7}{4}\)

\(M_{min}=\dfrac{7}{4}\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{2};\dfrac{1}{\sqrt{2}};1\right)\)

 Nguyễn Việt Lâm anh oiiiiiiiiiii

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)