\(A=\dfrac{x^2+y^2}{xy}+\dfrac{2xy}{x^2+y^2}=\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}+\dfrac{2xy}{x^2+y^2}\)

\(A\ge\dfrac{2xy}{2xy}+2\sqrt{\left(\dfrac{x^2+y^2}{2xy}\right)\left(\dfrac{2xy}{x^2+y^2}\right)}=3\)

Dấu "=" xảy ra khi \(x=y\)

\(B=\dfrac{\left(x+y\right)^2-4xy}{xy}+\dfrac{4xy}{\left(x+y\right)^2}=\dfrac{\left(x+y\right)^2}{xy}+\dfrac{4xy}{\left(x+y\right)^2}-4\)

\(B=\dfrac{\left(x+y\right)^2}{4xy}+\dfrac{4xy}{\left(x+y\right)^2}+\dfrac{3}{4}.\dfrac{\left(x+y\right)^2}{xy}-4\)

\(B\ge2\sqrt{\dfrac{\left(x+y\right)^2.4xy}{4xy.\left(x+y\right)^2}}+\dfrac{3}{4}.\dfrac{4xy}{xy}-4=1\)

\(B_{min}=1\) khi \(x=y\)

\(A=\left(x-y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(A_{min}=\dfrac{3}{4}\) khi \(x-y+\dfrac{1}{2}=0\)

\(B=\dfrac{7}{-\left(x-5\right)^2-5}\ge-\dfrac{7}{5}\)

\(B_{min}=-\dfrac{7}{5}\) khi \(x=5\)

Ta có: \(\left(x-1\right)^2+\left(x+y\right)^2\le9\Rightarrow x+y\le3\).

Áp dụng bất đẳng thức AM - GM ta có:

\(\dfrac{2}{x}+2x\ge2\sqrt{\dfrac{2}{x}.2x}=4;\dfrac{4}{y}+y\ge2\sqrt{\dfrac{4}{y}.y}=4\).

Do đó \(\dfrac{2}{x}\ge4-2x;\dfrac{4}{y}\ge4-y\)

\(\Rightarrow P\ge8-4\left(x+y\right)\ge-4\). (do \(x+y\le3\)).

Vậy...

Đẳng thức xảy ra khi và chỉ khi x = 1; y = 2.

\(A=\dfrac{-\left(x^2+2xy+y^2\right)+4x^2+4xy+y^2}{x^2+2xy+y^2}=-1+\left(\dfrac{2x+y}{x+y}\right)^2\ge-1\)

\(A_{min}=-1\) khi \(2x+y=0\)

Answer:

3.

\(x^2+2y^2+2xy+7x+7y+10=0\)

\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)

\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)

\(\Rightarrow4S^2+28S+4y^2+40=0\)

\(\Rightarrow4S^2+28S+49+4y^2-9=0\)

\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)

\(\Rightarrow-3\le2S+7\le3\)

\(\Rightarrow-10\le2S\le-4\)

\(\Rightarrow-5\le S\le-2\left(2\right)\)

Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)

Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)

Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)

Lời giải:
\(P=\sum \frac{1}{2xy^2+1}=\sum (1-\frac{2xy^2}{2xy^2+1})\)

\(=3-2\sum\frac{xy^2}{2xy^2+1}\geq 3-2\sum \frac{xy^2}{3\sqrt[3]{x^2y^4}}\) theo BĐT AM-GM.

\(=3-\frac{2}{3}\sum \sqrt[3]{xy^2}\)

Tiếp tục áp dụng BĐT AM-GM:

\(\sqrt[3]{xy^2}\leq \frac{x+y+y}{3}\Rightarrow \sum \sqrt[3]{xy^2}\leq \frac{3(x+y+z)}{3}=3\)

$\Rightarrow P\geq 3-\frac{2}{3}.3=1$

Vậy $P_{\min}=1$. Giá trị này đạt tại $x=y=z=1$

Lời giải:

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)(x^2+2yz+y^2+2xz+z^2+2xy)\geq (x+y+z)^2\)

\(\Leftrightarrow P(x+y+z)^2\geq (x+y+z)^2\)

\(\Rightarrow P\geq 1\)

Vậy \(P_{\min}=1\)

Dấu bằng xảy ra khi \(x=y=z\)

\(P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\)

Áp dụng BDT Cô-si : \(a^2+b^2\ge2ab\)

\(\Rightarrow\left\{{}\begin{matrix}y^2+z^2\ge2yz\\x^2+z^2\ge2xz\\x^2+y^2\ge2xy\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2\ge x^2+2yz>0\\x^2+y^2+z^2\ge y^2+2xz>0\\x^2+y^2+z^2\ge z^2+2xy>0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{x^2+y^2+z^2}\le\dfrac{x^2}{x^2+2yz}\\\dfrac{y^2}{x^2+y^2+z^2}\le\dfrac{y^2}{y^2+2xz}\\\dfrac{z^2}{x^2+y^2+z^2}\le\dfrac{z^2}{z^2+2xy}\end{matrix}\right.\\ \Rightarrow P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\\ \ge\dfrac{x^2}{x^2+y^2+z^2}+\dfrac{y^2}{x^2+y^2+z^2}+\dfrac{z^2}{x^2+y^2+z^2}\\ \ge\dfrac{x^2+y^2+z^2}{x^2+y^2+z^2}\ge1\forall x;y;z\)

Dấu "=" xảy ra khi \(:\left\{{}\begin{matrix}y=z\\x=z\\x=y\end{matrix}\right.\Leftrightarrow x=y=z\)

Vậy \(P_{Min}=1\) khi \(x=y=z\)