Đáp án:

Giải thích các bước giải:

 Sửa đề : `a,b,c>0`

Ta có

`3/2>= a+b+c`

`=>3>=2(a+b+c)`

Do `a,b,c>0`

`=>a+b+c>=3\root{3}{abc}`

`=>2(a+b+c)>=6\root{3}{abc}`

`=>3>=6\root{3}{abc}`

`=>1/2>=\root{3}{abc}`

`=>abc<=1/8`

`P=(3+1/a+1/b)(3+1/b+1/c)(3+1/c+1/a)`

`=>8P=(6+2/a+2/b)(6+2/b+2/c)(6+2/c+2/a)`

Ta có `a,b,c>0`

`=>6+2/a+2/b=2+2+2+1/a+1/a+1/b+1/b>=7\root{7}{2.2.2. 1/a. 1/a. 1/b. 1/b}`

`=>6+2/a+2/b>=7\root{7}{8/(ab)^2}`

Chứng minh tương tự

`6+2/b+2/c>=7\root{7}{8/(bc)^2},6+2/c+2/a>=7\root{7}{8/(ca)^2}`

`=>8P>=7\root{7}{8/(ab)^2} . 7\root{7}{8/(bc)^2} . 7\root{7}{8/(ca)^2}`

`=>8P>=343\root{7}{8^3/(abc)^4} >=343\root{7}{8^3/(1/8)^4}`

`=>8P>=343\root{7}{8^7}`

`=>8P>=343.8`

`=>P>=343`

Dấu `=` xảy rra`<=>a=b=c=1/2`

Đáp án:

a+b+c$\leq$ $\frac{3}{2}$ 

=> 3≥2(a+b+c)

=> P=[3+$\frac{1}{a}$ +$\frac{1}{b}$ ][3+$\frac{1}{c}$ +$\frac{1}{b}$ ][3+$\frac{1}{a}$ +$\frac{1}{c}$ ]

áp dụng BĐT co si  3 số dương

a+b+c≥3∛abc

=> 3≥6∛abc

=> 1/2≥∛abc

=> 1/8≥abc

P=[3+$\frac{1}{a}$ +$\frac{1}{b}$ ][3+$\frac{1}{c}$ +$\frac{1}{b}$ ][3+$\frac{1}{a}$ +$\frac{1}{c}$ ]

8P=([6+$\frac{2}{a}$ +$\frac{2}{b}$ ][6+$\frac{2}{c}$ +$\frac{2}{b}$ ][6+$\frac{2}{a}$ +$\frac{2}{c}$ ]

ta có

6+$\frac{2}{a}$ +$\frac{2}{b}$ ≥2+2+2+$\frac{1}{a}$ +$\frac{1}{b}$$\frac{1}{a}$ +$\frac{1}{b}$

áp dụng BĐT co si

2+2+2+$\frac{1}{a}$ +$\frac{1}{b}$$\frac{1}{a}$ +$\frac{1}{b}$≥7$\sqrt[7]{\frac{8}{(ab)^2}}$ 

tương tự ta có

6+$\frac{2}{c}$ +$\frac{2}{b}$≥7$\sqrt[7]{\frac{8}{(cb)^2}}$ 

6+$\frac{2}{a}$ +$\frac{2}{c}$ ]≥7$\sqrt[7]{\frac{8}{(ac)^2}}$ 

=> 8P≥7$\sqrt[7]{\frac{8}{(ac)^2}}$ .$\sqrt[7]{\frac{8}{(cb)^2}}$ .7$\sqrt[7]{\frac{8}{(ab)^2}}$ 

=> 8P≥343.7$\sqrt[7]{\frac{8^3}{(abc)^2}}$ ≥343$\sqrt[7]{8^7}$ ( vì abc≤1/8 (cmt))

=> P≥343

dấu "=" xảy ra <=> a=b=c=1/2