cho 2 số a,b thỏa mãn a-b=1.cmr a^3-b^3-ab>=1/2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
- Với \(a+b\ge4\Rightarrow A\le0\)
- Với \(a+b< 4\Rightarrow4-a-b>0\)
\(\Rightarrow A=\dfrac{a}{2}.\dfrac{a}{2}.b.\left(4-a-b\right)\)
\(\Rightarrow A\le\dfrac{1}{64}\left(\dfrac{a}{2}+\dfrac{a}{2}+b+4-a-b\right)^4=4\)
\(A_{max}=4\) khi \(\left(a;b\right)=\left(2;1\right)\)
2.
\(P=a+\dfrac{1}{2}.a.2b\left(1+2c\right)\le a+\dfrac{a}{8}\left(2b+1+2c\right)^2\)
\(P\le a+\dfrac{a}{8}\left(7-2a\right)^2=\dfrac{1}{8}\left(4a^3-28a^2+57a-36\right)+\dfrac{9}{2}\)
\(P\le\dfrac{1}{8}\left(a-4\right)\left(2a-3\right)^2+\dfrac{9}{2}\le\dfrac{9}{2}\)
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(\dfrac{3}{2};1;\dfrac{1}{2}\right)\)
Câu 3 bạn xem lại đề, mình có thể chắc chắn với bạn là đề sai
Ví dụ bạn cho \(x=98,y=100\) thì vế trái chỉ lớn hơn 8 một chút
Đề đúng phải là: \(\left(x+y\right)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{16xy}{\left(x-y\right)^2}\ge12\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A = \(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}=0\)
= \(\dfrac{a-b}{c^2+ab+bc+ca}+\dfrac{b-c}{a^2+ab+bc+ca}+\dfrac{c-a}{b^2+ab+bc+ca}\)
= \(\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(c+a\right)}+\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}\)
= \(\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c+a\right)\left(c-a\right)}{\left(c+a\right)\left(b+c\right)\left(a+b\right)}\)
= \(\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
\(\dfrac{a-b}{1+c^2}+\dfrac{b-c}{1+a^2}+\dfrac{c-a}{1+b^2}\)
\(=\dfrac{a-b}{ab+bc+ca+c^2}+\dfrac{b-c}{ab+bc+ca+a^2}+\dfrac{c-a}{ab+bc+ca+b^2}\)
\(=\dfrac{a-b}{\left(c+a\right)\left(c+b\right)}+\dfrac{b-c}{\left(a+b\right)\left(a+c\right)}+\dfrac{c-a}{\left(b+a\right)\left(b+c\right)}\)
\(=\dfrac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^2+b^2\le1+ab\)
\(\Leftrightarrow a^2-ab+b^2\le1\)
\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\le a+b\)
\(\Leftrightarrow a^3+b^3\le a+b\)
\(\Leftrightarrow\left(a^3+b^3\right)\left(a^3+b^3\right)\le\left(a+b\right)\left(a^5+b^5\right)\) ( \(a^3+b^3=a^5+b^5\))
\(\Leftrightarrow a^6+2a^3b^3+b^6\le a^6+ab^5+a^5b+b^6\)
\(\Leftrightarrow a^5b+ab^5\ge2a^3b^3\)
\(\Leftrightarrow a^5b+ab^5-2a^3b^3\ge0\)
\(\Leftrightarrow ab\left(a^4-2a^2b^2+b^4\right)\ge0\)
\(\Leftrightarrow ab\left(a^2-b^2\right)^2\ge0\) (luôn đúng \(\forall a;b>0\))
Vậy \(a^2+b^2\le1+ab\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
Áp dụng BĐT Bunhiacopxky ta có:
$(a^2+b^2+c^2)(1+1+1)\geq (a+b+c)^2$
$\Leftrightarrow 3(a^2+b^2+c^2)\geq 1$
$\Leftrightarrow a^2+b^2+c^2\geq \frac{1}{3}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c=\frac{1}{3}$
Bài 2:
Áp dụng BĐT Bunhiacopxky:
$(a^2+4b^2+9c^2)(1+\frac{1}{4}+\frac{1}{9})\geq (a+b+c)^2$
$\Leftrightarrow 2015.\frac{49}{36}\geq (a+b+c)^2$
$\Leftrightarrow \frac{98735}{36}\geq (a+b+c)^2$
$\Rightarrow a+b+c\leq \frac{7\sqrt{2015}}{6}$ chứ không phải $\frac{\sqrt{14}}{6}$ :''>>
a^3-b^3-ab>=1/2
(a-b)(a^2+b^2+ab)-ab>=1/2
a^2+b^2>=1/2 (vì a-b=1)
(a-b)^2+2ab-1/2>=0
1/2+2ab>=0 (vì a-b=1)
1+4ab>=0
(a-b)^2+4ab>=0
(a+b)^2>=0 (luôn đúng)