1 thách dám tích

\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}+\frac{1}{2xy}\\ =\frac{1}{4}+\frac{1}{2xy}\ge\frac{1}{4}+\frac{1}{8}=\frac{3}{8}\)

Dấu = xảy ra khi x=y=2

By Titu's Lemma we easy have:

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

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

\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)

\(=\frac{17}{4}\)

Mk xin b2 nha!

\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)

\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)

\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)

\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)

Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)

BĐT Bunhiacopxky em chưa học cô ạ

Cô cong cách nào không ạ

Nguyễn Thị Nguyệt Ánh:

Vậy thì bạn có thể chứng minh $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$ thông qua BĐT Cô-si:

Áp dụng BĐT Cô-si:

$x+y+z\geq 3\sqrt[3]{xyz}$

$xy+yz+xz\geq 3\sqrt[3]{x^2y^2z^2}$

Nhân theo vế:

$(x+y+z)(xy+yz+xz)\geq 9xyz$

$\Rightarrow \frac{xy+yz+xz}{xyz}\geq \frac{9}{x+y+z}$
hay $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$

\(xy+yz+zx=4xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)

Ta có \(M=\frac{1}{4\left(x+y\right)}+\frac{1}{4\left(y+z\right)}+\frac{1}{4\left(z+x\right)}\)

               \(=\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{y+z}+\frac{4}{z+x}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}\right)\)

                                                                                     \(=\frac{1}{8}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{8}.4=\frac{1}{2}\)

Dấu "=" tại x = y = z = ³/₄

\(\frac{1}{x^2}+\frac{1}{9y^2}\ge2\sqrt{\frac{1}{x^2}.\frac{1}{9y^2}}=\frac{2}{3xy}=\frac{2}{3}\)

Dấu \(=\)xảy ra khi \(\hept{\begin{cases}\frac{1}{x^2}=\frac{1}{9y^2}\\xy=1\end{cases}}\Rightarrow\hept{\begin{cases}x=\sqrt{3}\\y=\frac{1}{\sqrt{3}}\end{cases}}\).

\(2=\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\)

\(\Leftrightarrow xy\ge1\)

\(\Rightarrow x+y\ge2\sqrt{xy}\ge2\)