=x^2-6x+9+4y^2-8y+4+2010

=(x-3)^2+(2y-2)^2+2010>=2010

Dấu = xảy ra khi x=3 và y=1

Áp dụng bđt bunhia có:

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

\(\Leftrightarrow\dfrac{25}{4}\ge\left(x+y\right)^2\)\(\Leftrightarrow x+y\le\dfrac{5}{2}\)

Dấu = xảy ra\(\Leftrightarrow\left\{{}\begin{matrix}x=4y\\x^2+4y^2=5\end{matrix}\right.\Leftrightarrow\) \(\left\{{}\begin{matrix}16y^2+4y^2=5\\x=4y\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Ta có:

\(3-S=\left(x^2+4y^2+9z^2\right)-\left(2x+4y+6z\right)\)

\(\Leftrightarrow3-S=\left(x^2-2x+1\right)+\left(4y^2-4y+1\right)+\left(9z^2-6z+1\right)-3\)

\(\Leftrightarrow6-S=\left(x-1\right)^2+\left(2y-1\right)^2+\left(3z-1\right)^2\ge0\)

\(\Leftrightarrow S\le6\)

\(S_{max}=6\) khi \(\left\{{}\begin{matrix}x-1=0\\2y-1=0\\3z-1=0\end{matrix}\right.\) \(\Leftrightarrow\left(x;y;z\right)=\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\)

a) Áp dụng bất đẳng thức Cosi ta có :

\(x^2+1\geq 2x\\ 4y^2+1\geq 4y\\ 9z^2+1\geq 6z\)

Suy ra \(S\leq 6\)

Dấu = xảy ra khi \(x=1;y=\frac{1}{2}; z=\frac{1}{3}\)

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

x^2+4y^2+z^2-2x-6z+8y+15

=x^2+4y^2+z^2-2x-6z+8y+1+1+4+9

=(x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)+1

=(x-1)^2+4(y+1)^2+(z-3^)2+1

Ta thấy:(x−1)^2≥0

              4(y+1)^2≥0

             (z−3)^ 2≥0

{(x−1)^24(y+1)^2(z−3)^2≥0

⇒(x−1)^2+4(y+1)^2+(z−3)^2≥0

⇒(x−1)2+4(y+1)2+(z−3)2+1≥0+1=1>0

\(x^2+xy+y^2+1.=x^2+2.x.\dfrac{y}{2}+\left(\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1.\\ =\left(x+\dfrac{y}{2}\right)^2+\dfrac{3}{4}y^2+1>0\forall x;y\in R.\\ \Rightarrow x^2+xy+y^2+10\forall x;y\in R.\)

Kkk

\(x^2+4y^2+z^2-2x-6z+8y+14=0\\\Leftrightarrow (x^2-2x+1)+(4y^2+8y+4)+(z^2-6z+9)=0\\\Leftrightarrow (x^2-2\cdot x\cdot1+1^2)+[(2y)^2+2\cdot2y\cdot 2+2^2]+(z^2-2\cdot z\cdot3+3^2)=0\\\Leftrightarrow (x-1)^2+(2y+2)^2+(z-3)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-1\right)^2\ge0\forall x\\\left(2y+2\right)^2\ge0\forall y\\\left(z-3\right)^2\ge0\forall z\end{matrix}\right.\)

\(\Rightarrow\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2\ge0\forall x;y;z\)

Mặt khác: \(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\)

nên ta được: 

\(\left\{{}\begin{matrix}x-1=0\\2y+2=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\\z=3\end{matrix}\right.\)

Vậy: ...

\(x^2+4y^2+z^2-2x-6z+8y+14=0\)

\(\left(x^2-2x+1\right)+\left(4y^2+8y+4\right)+\left(z^2-6z+9\right)=0\)

\(\left(x-1\right)^2+\left(2y+2\right)^2+\left(z-3\right)^2=0\) (1)

Do \(\left(x-1\right)^2\ge0;\left(2y+2\right)^2\ge0;\left(z-3\right)^2\ge0\)

\(\left(1\right)\Rightarrow\) \(\left(x-1\right)^2=0;\left(2y+2\right)^2=0;\left(z-3\right)^2=0\)

*) \(\left(x-1\right)^2=0\)

\(x-1=0\)

\(x=1\)

*) \(\left(2y+2\right)^2=0\)

\(2y+2=0\)

\(2y=-2\)

\(y=-1\)

*) \(\left(z-3\right)^2=0\)

\(z-3=0\)

\(z=3\)

Vậy x = 1; y = -1; z = 3

Latex của mình có chút lỗi nên xin phép đánh máy hoàn toàn nhé
x^2+4y^2+z^2-2x-6z+8y+14=0
<=> (x^2-2x+1) +4(y^2+2y+1) +(z^2-6z+9)=0
<=> (x-1)^2+4(y+1)^2+(z-3)^2=0
Do (x-1)^2, (y+1)^2>=0, (z-3)^2>=0
Nên x-1=0, y+1=0, z-3=0
<=> x=1, y=-1, z=3