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Giải thích các bước giải:

a.Ta có:

$3(a^2+b^2+c^2)=(1^2+1^2+1^2)(a^2+b^2+c^2)\ge (1\cdot a+1\cdot b+1\cdot c)^2=(a+b+c)^2$

b.Ta có:

$\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac1{z}$ 

$=(\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac1{z})\cdot 1$

$=(\dfrac{1}{16x}+\dfrac{1}{4y}+\dfrac1{z})\cdot (x+y+z)$

$\ge(\sqrt{\dfrac1{16x}\cdot x}+\sqrt{\dfrac1{4y}\cdot y}+\sqrt{\dfrac1z\cdot z})^2=\dfrac{49}{16}$

Dấu = xảy ra khi:

$\dfrac{1}{\sqrt{16x}}:\sqrt{x}=\dfrac{1}{\sqrt{4y}}:\sqrt{y}=\dfrac{1}{\sqrt{z}}:\sqrt{z}$

$\to \dfrac1{4x}=\dfrac{1}{2y}=\dfrac{1}z$

$\to 4x=2y=z$

$\to x=\dfrac14z, y=\dfrac12z$

Mà $x+y+z=1$

$\to \dfrac14z+\dfrac12z+z=1$

$\to z=\dfrac47$

$\to x=\dfrac17, y=\dfrac27$