Cho các số dương a,b,c thỏa mãn a+b+c=3 Chứng minh rằng: $\frac{bc}{2a+b+c}$+$\frac{ac}{2b+a+c}$+$\frac{ba}{2c+a+b}$$\leq$ $\frac{3}{4}$

Ta có:

`4((bc)/(2a + b + c) + (ac)/(2b + a + c) + (ab)/(2c + a + b))`

`= (4bc)/(2a + b + c) + (4ac)/(2b + a + c) + (4ab)/(2c + a + b)`

`= bc . 4/((a + b) + (a + c)) + ac . 4/((a + b) + (b + c)) + ab . 4/((a + c) + (b + c))`

Áp dụng BĐT `4/(x + y) <= 1/x + 1/y`, ta có:

$\bullet$ `4/((a + b) + (a + c)) ≤ 1/(a + b) + 1/(a + c)`

`⇔ bc . 4/((a + b) + (a + c)) ≤ bc(1/(a + b) + 1/(a + c))`

$\bullet$ `4/((a + b) + (b + c)) ≤ 1/(a + b) + 1/(b + c)`

`⇔ ac . 4/((a + b) + (b + c)) ≤ ac(1/(a + b) + 1/(b + c))`

$\bullet$ `4/((a + c) + (b + c)) ≤ 1/(a + c) + 1/(b + c)`

`⇔ ab . 4/((a + c) + (b + c)) ≤ ab(1/(a + c) + 1/(b + c))`

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Ta có:

`bc(1/(a + b) + 1/(a + c)) + ac(1/(a + b) + 1/(b + c)) + ab(1/(a + c) + 1/(b + c))`

`= (bc)/(a + b) + (bc)/(a + c) + (ac)/(a + b) + (ac)/(b + c) + (ab)/(a + c) + (ab)/(b + c)`

`= (ac + bc)/(a + b) + (ab + bc)/(a + c) + (ac + ab)/(b + c)`

`= [c(b + a)]/(a + b) + [b(a + c)]/(a + c) + [a(b + c)]/(c + b)`

`= a + b + c`

`= 3`

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`⇒ bc . 4/((a + b) + (a + c)) + ac . 4/((a + b) + (b + c)) + ab . 4/((a + c) + (b + c))`

`≤ bc(1/(a + b) + 1/(a + c)) + ac(1/(a + b) + 1/(b + c)) + ab(1/(a + c) + 1/(b + c)) = 3`

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`⇒ 4((bc)/(2a + b + c) + (ac)/(2b + a + c) + (ab)/(2c + a + b)) ≤ 3`

`⇔ (bc)/(2a + b + c) + (ac)/(2b + a + c) + (ab)/(2c + a + b) ≤ 3/4`

Dấu $``$$="$ xảy ra khi: `a = b = 1`