bất đẳng thức

J

jet_nguyen

Lời giải: Dễ thấy: $$A=\left(16.\left(y+\dfrac{x}{2}\right)^2+\dfrac{9z^2}{4}\right)+\left(12x^2+3.\left( y + \dfrac{z}{2} \right)^2 \right) .$$$$=16.\left(x^2+xy+y^2\right)+3.\left(y^2+yz+z^2 \right)=96$$
Ta có bất đẳng thức $$u^2+v^2\ge 2uv$$ Vậy:
$$A\ge 12.\left(y+\frac{x}{2}\right).z+12.x.\left(y+\frac{z}{2}\right)=12.\left(xy+yz+zx\right).$$
 
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