Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{x^3+y^3+xy}+\frac{4x^2y^2+2}{xy}=\frac{1}{\left(x+y\right)\left(x^2-xy+y^2\right)+xy}+4xy+\frac{2}{xy}\)
\(=\frac{1}{x^2+y^2}+4xy+\frac{2}{xy}\)
\(=\left(4xy+\frac{1}{4xy}\right)+\left(\frac{7}{4xy}+\frac{1}{x^2+y^2}\right)\)
\(=\left(4xy+\frac{1}{4xy}\right)+\left(\frac{1}{2xy}+\frac{1}{x^2+y^2}\right)+\frac{5}{4xy}\)
\(\ge2\sqrt{4xy.\frac{1}{4xy}}+\frac{4}{x^2+y^2+2xy}+\frac{5}{\left(x+y\right)^2}=5+4+2=11\)
Dấu "=" khi x=y=1/2
Áp dụng BĐT svacxơ, ta có
\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\ge4\)
Dấu = xảy ra <=>x=y=1/2
^_^
Bài 1:
\(\frac{2}{x^2+2y^2+3}=\frac{2}{\left(x^2+y^2\right)+\left(y^2+1\right)+2}\le\frac{2}{2xy+2y+2}=\frac{1}{xy+y+1}\)
Bài 2:
\(A=\frac{4}{4x^2+9y^2}+\frac{4}{12xy}+\frac{52}{2x.3y}\ge\frac{16}{4x^2+9y^2+12xy}+\frac{52.4}{\left(2x+3y\right)^2}\)
\(A\ge\frac{16}{\left(2x+3y\right)^2}+\frac{208}{\left(2x+3y\right)^2}=\frac{224}{\left(2x+3y\right)^2}\ge\frac{224}{4}=56\)
\(A_{min}=56\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{3}\end{matrix}\right.\)
ta có: \(\frac{\sqrt{2x^2+y^2}}{xy}=\sqrt{\frac{2}{y^2}+\frac{1}{x^2}}\)
Áp dụng BĐT bunyakovsky:\(\left(2+1\right)\left(\frac{2}{y^2}+\frac{1}{x^2}\right)\ge\left(\frac{2}{y}+\frac{1}{x}\right)^2\)
\(\Rightarrow\frac{2}{y^2}+\frac{1}{x^2}\ge\frac{1}{3}\left(\frac{2}{y}+\frac{1}{x}\right)^2\).....bla bla
áp dụng bdt cauchy -schửat dạng engel ta có
\(A=\frac{x^2}{x+y}+\frac{y^2}{y+z}+\frac{z^2}{x+z}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{x+y+z}{2}\)\(\ge\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2}=\frac{1}{2}\)
(do \(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\) bn tự cm nhé)
dau = xay ra \(\Leftrightarrow x=y=z=\frac{1}{3}\)
\(A=\frac{1}{x^2+xy+y^2}+\frac{\frac{1}{9}}{xy}+4xy+\frac{1}{4xy}+\frac{23}{36xy}\)
\(A\ge\frac{\left(1+\frac{1}{3}\right)^2}{x^2+2xy+y^2}+2\sqrt{\frac{4xy}{4xy}}+\frac{23}{9\left(x+y\right)^2}\)
\(A\ge\frac{16}{9\left(x+y\right)^2}+2+\frac{23}{9\left(x+y\right)^2}=\frac{19}{3}\)
\(A_{min}=\frac{19}{3}\) khi \(x=y=\frac{1}{2}\)