\(P=x^2+3x+y^2+3y+\frac{9}{x^2+y^2+1}\)

\(=x^2+y^2+1+\frac{9}{x^2+y^2+1}+3x+3y-1\)

\(\ge2.3.\frac{\sqrt{x^2+y^2+1}}{\sqrt{x^2+y^2+1}}+2.3.\sqrt{xy}-1\)

\(=6+6-1=11\)

Dấu = xảy ra khi x = y = 1

1) \(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)\(\Leftrightarrow\)\(x+y\ge8\)

\(\frac{1}{2}=\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\)\(\Leftrightarrow\)\(xy=2\left(x+y\right)\ge16\)

\(A=\sqrt{x}+\sqrt{y}\ge2\sqrt[4]{xy}\ge2\sqrt[4]{16}=4\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=4\)

2) \(B=\sqrt{3x-5}+\sqrt{7-3x}\ge\sqrt{3x-5+7-3x}=\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{7}{3}\end{cases}}\)

\(B=\sqrt{3x-5}+\sqrt{7-3x}\le\frac{3x-5+1+7-3x+1}{2}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=2\)

Áp dụng bất đẳng thức svác sơ ta có 

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

Đặt \(P=\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\)

Áp dụng bất đẳng thức Canchy Schwarz dạng Engel : 

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

Dấu " = " xảy ra khi x=y=z=1.

Sử dụng AM - GM ta dễ có:

\(\frac{x^2}{y+3z}+\frac{y+3z}{16}\ge2\sqrt{\frac{x^2}{y+3z}\cdot\frac{y+3z}{16}}=\frac{x}{2}\)

Tương tự:

\(\frac{y^2}{z+3x}+\frac{z+3x}{16}\ge\frac{y}{2};\frac{z^2}{x+3y}+\frac{x+3y}{16}\ge\frac{z}{2}\)

Khi đó:

\(\frac{x^2}{y+3z}+\frac{y^2}{z+3x}+\frac{z^2}{x+3y}\ge\frac{x+y+z}{2}-\frac{x+y+z}{4}=\frac{x+y+z}{4}=\frac{3}{4}\)

Đẳng thức xảy ra tại x=y=z=1

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

Ta có:\(3x^2-4xy+3y^2=25\)

\(\Leftrightarrow2x^2-4xy+2y^2+x^2+y^2=25\)

\(\Leftrightarrow2\left(x-y\right)^2+x^2+y^2=25\Leftrightarrow x^2+y^2=25-2\left(x-y\right)^2\le25\)

\(\Rightarrow\)GTLN của P là 25 đạt được khi x=y\(\Rightarrow3x^2-4x^2+3x^2=25\Rightarrow2x^2=25\Rightarrow x=\frac{5}{\sqrt{2}}=y\)

Lại có:\(3x^2-4xy+3y^2=25\Leftrightarrow3\left(x^2+y^2\right)=25+4xy\)

\(\Leftrightarrow3\left(x^2+y^2\right)+2\left(x^2+y^2\right)=25+2x^2+4xy+2y^2\)

\(\Leftrightarrow5\left(x^2+y^2\right)=25+2\left(x+y\right)^2\ge25\)

\(\Rightarrow x^2+y^2\ge5\)

\(\Rightarrow\)GTNN của P là 5 đạt được khi \(x=-y\Rightarrow3x^2+4x^2+3x^2=25\Rightarrow10x^2=25\Rightarrow x^2=\frac{5}{2}\Rightarrow x=\sqrt{\frac{5}{2}}\)

 \(\Rightarrow y=-\sqrt{\frac{5}{2}}\)