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

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

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

Giả thiết tương đương \(\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=29\).

Áp dụng bđt Cauchy - Schwarz ta có:

\(\left(2x-3y+4z-20\right)^2=\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]=29^2\Rightarrow\left|2x-3y+4z-20\right|\le29\)

điều kiện ban đầu <=> (x-1)²+(y-2)²+(z-3)² \(\le1\)

áp dụng bdt sau (ax+ by+ cz)²\(\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)(bunhiacopxky với 3 số)

[ x-1 + 2(y-2) + 2(z-3)]² \(\le\left(1^2+2^2+2^2\right)\left[\left(x-1\right)^2+\left(y-2\right)^2+\left(z-2\right)^2\right]\le9.1=9\)

=>\(-3\le\) x-1 +2(y-2) +2(z-3) \(\le3\) <=> 8\(\le x+2y+2z\le14\)

your gay

Lời giải:

$x^2+4y^2+9z^2=2x+4y+6z-3$

$\Leftrightarrow (x^2-2x+1)+(4y^2-4y+1)+(9z^2-6z+1)=0$

$\Leftrightarrow (x-1)^2+(2y-1)^2+(3z-1)^2=0$

Ta thấy: $(x-1)^2\geq 0; (2y-1)^2\geq 0; (3z-1)^2\geq 0$ với mọi $x,y,z\in\mathbb{R}$

Do đó để tổng của chúng bằng $0$ thì:

$(x-1)^2=(2y-1)^2=(3z-1)^2=0$

$\Leftrightarrow x=1; y=\frac{1}{2}; z=\frac{1}{3}$
Khi đó:

$xyz=1.\frac{1}{2}.\frac{1}{3}=\frac{1}{6}$