- Bổ sung đề: CMR \(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge6\).

- Ta có: \(\dfrac{1+3x}{1+y^2}=\left(1+3x\right)-\dfrac{y^2\left(1+3x\right)}{1+y^2}\ge\left(1+3x\right)-\dfrac{y^2\left(1+3x\right)}{2y}=1+3x-\dfrac{y\left(1+3x\right)}{2}=1+3x-\dfrac{y}{2}-\dfrac{3xy}{2}\left(1\right)\)

- Tương tự, ta cũng có:

\(\dfrac{1+3y}{1+z^2}=1+3y-\dfrac{z}{2}-\dfrac{3yz}{2}\left(2\right)\), \(\dfrac{1+3z}{1+x^2}=1+3z-\dfrac{x}{2}-\dfrac{3zx}{2}\left(3\right)\)

- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\), ta được:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(xy+yz+zx\right)=3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}.3=\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(4\right)\)

- Mặt khác: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3.3=9\Rightarrow x+y+z\ge3\left(5\right)\)

- Từ (4), (5) ta có:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge\dfrac{5}{2}.3-\dfrac{3}{2}=6\left(đpcm\right)\)

- Dấu "=" xảy ra khi \(x=y=z=1\)

=>căn (4x+8)=4

=>4x+8=16

=>4x=8

hay x=2

\(\Leftrightarrow\sqrt{4x+8}=4\Leftrightarrow4x+8=16\Leftrightarrow4x=8\Leftrightarrow x=2\)

=>16x-2=78

=>16x=80

hay x=5

`<=> 16x - 2 = 78`

`<=> 16x = 80`

`-> x = 5`

1+2+3+4+5+6+7+8+9+10=5555

1+2+3+4+5+6+7+8+9+10=5555

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