(a+b+c) ³ ≥ a³+b³+c³+24abc với a,b,c ≥0 Giúp mik với ạ 25/08/2021 Bởi Hadley (a+b+c) ³ ≥ a³+b³+c³+24abc với a,b,c ≥0 Giúp mik với ạ
$\begin{cases} (\sqrt a – \sqrt b)^2 \geq 0 \\(\sqrt b – \sqrt c)^2 \geq 0 \\ (\sqrt a – \sqrt c)^2 \geq 0\end{cases} (a,b,c\geq0)$ ⇔ $\begin{cases} a + b \geq 2\sqrt{ab} \\ b + c \geq 2\sqrt{bc} \\ a + c \geq 2\sqrt{ac}\end{cases}$ Suy ra: $(a+b)(b+c)(a+c) \geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac})$ ⇔ $3(a+b)(b+c)(a+c) \geq 3(2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac})$ ⇔ $3(a+b)(b+c)(a+c) \geq 24abc$ ⇔ $a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) \geq a^3 + b^3 + c^3 + 24abc$ ⇔ $(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc$ Bình luận
Giả sử: `(a+b+c)^3≥ a^3+b^3+c^3+24abc` `⇔ (a+b+c)^3 -(a^3+b^3+c^3+24abc)≥0` `⇔ (a+b+c)^3 -a^3-b^3-c^3-24abc≥0` Bây giờ ta sẽ đi chứng minh ` (a+b+c)^3 -a^3-b^3-c^3-24abc≥0` Ta có: `(a+b+c)^3=[(a+b)+c]^3=(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3` `=a^3+3a^2b+3ab^2+b^3+3c(a^2+2ab+b^2)+3ac^2+3bc^2+c^3` `=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3.` Ta thay: `(a+b+c)^3=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3` vào biểu thưc cần chứng minh thì ta có: `=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3-a^3-b^3-c^3-24abc` `=(a^3-a^3)+(b^3-b^3)+(c^3-c^3)+(6abc-24abc)+3a^2b+3ab^2+3ca^2+3cb^2+3ac^2+3bc^2` `= 3a^2b+3ab^2+3ca^2+3cb^2+3ac^2+3bc^2-18abc` `=(3a^2b-6abc+3bc^2)+(3ab^2+3ac^2-6abc)+(3a^2c+3b^2c-6abc)` `=3b(a^2-2ac+c^2)+3a(b^2+c^2-2bc)+3c(a^2-2ab+b^2)` `=3b(a-c)^2+3a(b-c)^2+3c(a-b)^2≥0` (luôn đúng) Dấu bằng xảy ra khi `a=b=c.` Vậy `(a+b+c)^3≥ a^3+b^3+c^3+24abc.` Bình luận
$\begin{cases} (\sqrt a – \sqrt b)^2 \geq 0 \\(\sqrt b – \sqrt c)^2 \geq 0 \\ (\sqrt a – \sqrt c)^2 \geq 0\end{cases} (a,b,c\geq0)$
⇔ $\begin{cases} a + b \geq 2\sqrt{ab} \\ b + c \geq 2\sqrt{bc} \\ a + c \geq 2\sqrt{ac}\end{cases}$
Suy ra: $(a+b)(b+c)(a+c) \geq (2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac})$
⇔ $3(a+b)(b+c)(a+c) \geq 3(2\sqrt{ab})(2\sqrt{bc})(2\sqrt{ac})$
⇔ $3(a+b)(b+c)(a+c) \geq 24abc$
⇔ $a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) \geq a^3 + b^3 + c^3 + 24abc$
⇔ $(a+b+c)^3 \geq a^3 + b^3 + c^3 + 24abc$
Giả sử: `(a+b+c)^3≥ a^3+b^3+c^3+24abc`
`⇔ (a+b+c)^3 -(a^3+b^3+c^3+24abc)≥0`
`⇔ (a+b+c)^3 -a^3-b^3-c^3-24abc≥0`
Bây giờ ta sẽ đi chứng minh ` (a+b+c)^3 -a^3-b^3-c^3-24abc≥0`
Ta có: `(a+b+c)^3=[(a+b)+c]^3=(a+b)^3+3(a+b)^2c+3(a+b)c^2+c^3`
`=a^3+3a^2b+3ab^2+b^3+3c(a^2+2ab+b^2)+3ac^2+3bc^2+c^3`
`=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3.`
Ta thay: `(a+b+c)^3=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3` vào biểu thưc cần chứng minh thì ta có:
`=a^3+3a^2b+3ab^2+b^3+3ca^2+6cab+3cb^2+3ac^2+3bc^2+c^3-a^3-b^3-c^3-24abc`
`=(a^3-a^3)+(b^3-b^3)+(c^3-c^3)+(6abc-24abc)+3a^2b+3ab^2+3ca^2+3cb^2+3ac^2+3bc^2`
`= 3a^2b+3ab^2+3ca^2+3cb^2+3ac^2+3bc^2-18abc`
`=(3a^2b-6abc+3bc^2)+(3ab^2+3ac^2-6abc)+(3a^2c+3b^2c-6abc)`
`=3b(a^2-2ac+c^2)+3a(b^2+c^2-2bc)+3c(a^2-2ab+b^2)`
`=3b(a-c)^2+3a(b-c)^2+3c(a-b)^2≥0` (luôn đúng)
Dấu bằng xảy ra khi `a=b=c.`
Vậy `(a+b+c)^3≥ a^3+b^3+c^3+24abc.`