\(P=3x+\dfrac{12}{x}+y+\dfrac{16}{y}+2\left(x+y\right)\ge2\sqrt{3x.\dfrac{12}{x}}+2\sqrt{y.\dfrac{16}{y}}+2.6=32\)

\(\Rightarrow P_{min}=32\) khi \(\left\{{}\begin{matrix}3x=\dfrac{12}{x}\\y=\dfrac{16}{y}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Áp dụng bđt Cauchy Schwarz dạng Engel:

P=\(\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\ge\frac{\left(x+y+z\right)^2}{y+3z+z+3x+x+3y}=\frac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\frac{3^2}{4.3}=\frac{3}{4}\)

Dấu "=" xảy ra khi x=y=z=1

Lời giải:

Áp dụng BĐT AM-GM:
$1=x+y\geq 2\sqrt{xy}\Rightarrow xy\leq \frac{1}{4}$
$P=x^2y^2+\frac{1}{x^2y^2}+2-\frac{17}{6}$

$=x^2y^2+\frac{1}{x^2y^2}-\frac{5}{6}$

$=(x^2y^2+\frac{1}{256x^2y^2})+\frac{255}{256x^2y^2}-\frac{5}{6}$

$\geq 2\sqrt{\frac{1}{256}}+\frac{255}{256.\frac{1}{4^2}}-\frac{5}{6}=\frac{731}{48}$

Vậy $P_{\min}=\frac{731}{48}$ khi $x=y=\frac{1}{2}$

VV

CHÚC BẠN HỌC TỐT

Áp dụng BĐT Svac ta có:
\(P=\dfrac{x^2}{y+3z}+\dfrac{y^2}{z+3x}+\dfrac{z^2}{x+3y}\ge\dfrac{\left(x+y+z\right)^2}{4\left(x+y+z\right)}=\dfrac{x+y+z}{4}=\dfrac{3}{4}\)

Dấu '=' xảy ra khi \(x=y=z=1\)

Vậy \(P_{min}=\dfrac{3}{4}\) khi \(x=y=z=1\)

\(2P=2x^2+8y^2+\dfrac{150}{x}+\dfrac{2}{y}\)

\(=\dfrac{7}{5}x^2+7y^2+\left(\dfrac{3}{5}x^2+\dfrac{75}{x}+\dfrac{75}{x}\right)+\left(y^2+\dfrac{1}{y}+\dfrac{1}{y}\right)\) 

Ta có: \(\left(5+1\right)\left(x^2+5y^2\right)\ge5\left(x+y\right)^2\Rightarrow\dfrac{7\left(x^2+5y^2\right)}{5}\ge\dfrac{7\left(x+y\right)^2}{6}\ge42\)

\(\Rightarrow2P\ge42+3\sqrt[3]{\dfrac{3.75^2.x^2}{5x^2}}+3\sqrt[3]{\dfrac{y^2}{y^2}}=90\)

\(\Rightarrow P\ge45\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(5;1\right)\)

a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)

b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)

c) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5}{3}\left(x\ne y\right)\)

d) \(\dfrac{x^2-y^2}{x+y}=x-y\left(đk:x\ne-y\right)\)

e) \(\dfrac{x^3-x^2+x-1}{x^2-1}=\dfrac{x^2+1}{x+1}\left(đk:x\ne\pm1\right)\)

f) \(\dfrac{x^2+4x+4}{2x+4}=\dfrac{x+2}{2}\left(đk:x\ne-2\right)\)