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Ta có:
\(a^3+b^3\ge\frac{1}{4}\left(a+b\right)^3\)
\(\Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\)
\(\Leftrightarrow4a^3+4b^3\ge a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow3a^3+3b^3-3a^2b-3ab^2\ge0\)
\(\Leftrightarrow a^3-a^2b-ab^2+b^3\ge0\)( chia 2 vế cho 3)
\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2.\left(a+b\right)\ge0\)(Luôn đúng vì a,b>0)
\(\Rightarrowđpcm\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(a^4+b^4\ge a^3b+b^3a\)
\(\Leftrightarrow a^4-a^3b+b^4-b^3a\ge0\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
ta thấy : \(\orbr{\orbr{\begin{cases}\left(a-b\right)^2\ge0\\\left(a^2+ab+b^2\right)\ge0\end{cases}}}\Leftrightarrow dpcm\)
Dấu " = " xảy ra khi a = b
tk nka !!!! mk cố giải mấy bài nữa !11
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Lời giải:
a) Áp dụng BĐT Cô-si cho các số dương:
$a^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}a$
$b^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}b$
$\Rightarrow a^3+b^3+\frac{1}{2}\geq \frac{3}{4}(a+b)=\frac{3}{4}$
$\Rightarrow a^3+b^3\geq \frac{1}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^3+b^3}+\frac{3}{ab}=\frac{1}{a^2-ab+b^2}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}\geq \frac{(1+1+1+1)^2}{a^2-ab+b^2+ab+ab+ab}\)
\(=\frac{16}{(a+b)^2}=16\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
![](https://rs.olm.vn/images/avt/0.png?1311)
dễ mà cô nương
\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\left(a^2+ab+b^2\right)=\left\{\left(a+b\right)^2-ab\right\}\)
\(a^3-b^3=\left(a-b\right)\left(25-6\right)=19\left(a-b\right)\)
ta có
\(a=-5-b\)
suy ra
\(a^3-b^3=19\left(-5-2b\right)\) " xong "
2, trên mạng đầy
3, dytt mọe mày ngu ab=6 thì cmm nó phải chia hết cho 6 chứ :)
4 . \(x^2-\frac{2.1}{2}x+\frac{1}{4}+\frac{1}{3}-\frac{1}{4}>0\) tự làm dcmm
5. trên mạng đầy
6 , trên mang jđầy
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ a+b+c=6 \(\Rightarrow\)a+b=6-c
Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9
\(\Leftrightarrow\)ab=9-c(a+b)
Mà a+b=6-c (cmt)
\(\Rightarrow\)ab=9-c(6-c)
\(\Rightarrow\)ab=9-6c+c2
Ta có: (b-a)2\(\ge\)0 \(\forall\)b, c
\(\Rightarrow\)b2+a2-2ab\(\ge\)0
\(\Rightarrow\)(b+a)2-4ab\(\ge\)0
\(\Rightarrow\)(a+b)2\(\ge\)4ab
Mà a+b=6-c (cmt)
ab= 9-6c+c2 (cmt)
\(\Rightarrow\)(6-c)2\(\ge\)4(9-6c+c2)
\(\Rightarrow\)36+c2-12c\(\ge\)36-24c+4c2
\(\Rightarrow\)36+c2-12c-36+24c-4c2\(\ge\)0
\(\Rightarrow\)-3c2+12c\(\ge\)0
\(\Rightarrow\)3c2-12c\(\le\)0
\(\Rightarrow\)3c(c-4)\(\le\)0
\(\Rightarrow\)c(c-4)\(\le\)0
\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)
*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)
*
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1. (a+b)^2 ≥ 4ab
<=> a2+2ab+b2≥ 4ab
<=> a2+2ab+b2-4ab≥ 0
<=> a2-2ab+b2≥ 0
<=> (a-b)^2 ≥ 0 ( luôn đúng )
2. a^2 + b^2 + c^2 ≥ ab + bc + ca
<=> 2a^2 + 2b^2 + 2c^2 ≥ 2ab + 2bc + 2ca
<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0
<=> (a^2- 2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2) ≥ 0
<=> (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0 ( luôn đúng)
Xét hiệu \(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3\) ta có:
\(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3=\frac{1}{4}\left[4\left(a^3+b^3\right)-\left(a+b\right)^3\right]\)
\(=\frac{1}{4}\left[4a^3+4b^3-\left(a^3+3a^2b+3ab^2+b^3\right)\right]\)\(=\frac{1}{4}\left(4a^3+4b^3-a^3-3a^2b-3ab^2-b^3\right)\)
\(=\frac{1}{4}\left(3a^3+3b^3-3a^2b-3ab^2\right)\)\(=\frac{3}{4}\left(a^3+b^3-a^2b-ab^2\right)\)
\(=\frac{3}{4}\left[\left(a^3-a^2b\right)+\left(b^3-ab^2\right)\right]\)\(=\frac{3}{4}\left[a^2\left(a-b\right)+b^2\left(b-a\right)\right]\)
\(=\frac{3}{4}\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)\(=\frac{3}{4}\left(a-b\right)\left(a^2-b^2\right)\)\(=\frac{3}{4}\left(a-b\right)^2\left(a+b\right)\)
Vì a và b > 0 \(\Rightarrow a+b>0\)
mà \(\left(a-b\right)^2\ge0\forall a,b\)và \(\frac{3}{4}>0\)
\(\Rightarrow\frac{3}{4}\left(a-b\right)^2\left(a+b\right)\ge0\)
hay \(\left(a^3+b^3\right)-\frac{1}{4}\left(a+b\right)^3\ge0\)\(\Rightarrow a^3+b^3\ge\frac{1}{4}\left(a+b\right)^3\)