VP `=(a+b)(a^2-ab+b^2)`

`=a^3-a^2b+ab^2+a^2b-ab^2+b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)+b^3`

`=a^3+b^3`

.

VP `=(a-b)(a^2+ab+b^2)`

`=a^3+a^2b+ab^2-a^2b-ab^2-b^3`

`=a^3+(a^2b-a^2b)+(ab^2-ab^2)-b^3`

`=a^3-b^3`

đúng rồi mà

Ta có: \(a^3+b^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=\left(a+b\right)\left(a^2+2ab+b^2-3ab\right)\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)\)

Ta có: \(a^3-b^3\)

\(=\left(a-b\right)^3+3ab\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2-2ab+b^2+3ab\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2\right)\)

a )

`VP= (a+b)^3-3ab(a+b)`

     `=a^3+3a^2b+3ab^2+b^3-3a^2b-3ab^2`

     `=a^3+b^3 =VT (đpcm)`

b) 

b) Ta có

a) Ta có:

`VP= (a+b)^3-3ab(a+b)`

     `=a^3 + b^3+3ab ( a + b )- 3ab ( a + b )`

     `=a^3 + b^3=VT(dpcm)`

b) Ta có

Thực hiện phép nhân đa thức với đa thức ở vế trái. 

=> VT = VP (đpcm)

Câu 9:

\(a,\left(a+1\right)^2\ge4a\\ \Leftrightarrow a^2+2a+1\ge4a\\ \Leftrightarrow a^2-2a+1\ge0\\ \Leftrightarrow\left(a-1\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=1\)

\(b,\) Áp dụng BĐT cosi: \(\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge2\sqrt{a}\cdot2\sqrt{b}\cdot2\sqrt{c}=8\sqrt{abc}=8\)

Dấu \("="\Leftrightarrow a=b=c=1\)

Câu 10:

\(a,\left(a+b\right)^2\le2\left(a^2+b^2\right)\\ \Leftrightarrow a^2+2ab+b^2\le2a^2+2b^2\\ \Leftrightarrow a^2-2ab+b^2\ge0\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b\)

\(b,\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac\le3a^2+3b^2+3c^2\\ \Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(luôn.đúng\right)\)

Dấu \("="\Leftrightarrow a=b=c\)

Câu 13:

\(M=\left(a^2+ab+\dfrac{1}{4}b^2\right)-3\left(a+\dfrac{1}{2}b\right)+\dfrac{3}{4}b^2-\dfrac{3}{2}b+2021\\ M=\left[\left(a+\dfrac{1}{2}b\right)^2-2\cdot\dfrac{3}{2}\left(a+\dfrac{1}{2}b\right)+\dfrac{9}{4}\right]+\dfrac{3}{4}\left(b^2-2b+1\right)+2018\\ M=\left(a+\dfrac{1}{2}b-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\left(b-1\right)^2+2018\ge2018\\ M_{min}=2018\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{2}b=\dfrac{3}{2}\\b=1\end{matrix}\right.\Leftrightarrow a=b=1\)

Câu 6:

$2=(a+b)(a^2-ab+b^2)>0$

$\Rightarrow a+b>0$

$4(a^3+b^3)-N^3=4(a^3+b^3)-(a+b)^3$

$=3(a^3+b^3)-3ab(a+b)=(a+b)(a-b)^2\geq 0$
$\Rightarrow N^3\leq 4(a^3+b^3)=8$

$\Rightarrow N\leq 2$

Vậy $N_{\max}=2$

Câu 7:

BĐT $\Leftrightarrow a^3+b^3\geq ab(a+b)$

$\Leftrightarrow a^3+b^3-ab(a+b)\geq 0$

$\Leftrightarrow (a-b)^2(a+b)\geq 0$ (luôn đúng với mọi $a,b,c>0$)

Vậy ta có đpcm

Dấu "=" xảy ra khi $a=b>0$, $c$ dương bất kỳ. 

Biến đổi vế trái ta có:

VT =  a 3 + b 3 =(a+b)( a 2 -ab+ b 2 )

=(a+b)( a 2 -2ab+ b 2 +ab)

=(a + b)[ a - b 2  + ab] = VP

Vế phải bằng vế trái nên đẳng thức được chứng minh.

\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 5:

\(a+b=1\Rightarrow a=1-b\)

\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)

\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 7:

\(a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

giúp mình vs

5.

Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=a^2+b^2-ab\)

\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)

\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)

6.

Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)

Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)

Mặt khác với mọi a;b ta có:

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)

Từ đó:

\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)

\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)

\(N_{max}=2\) khi \(a=b=1\)

7.

Ta có:

\(a^3+b^3+abc=\left(a+b\right)\left(a^2+b^2-ab\right)+abc\ge\left(a+b\right)\left(2ab-ab\right)+abc\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

8.

\(\left|a+b\right|>\left|a-b\right|\Leftrightarrow\left(a+b\right)^2>\left(a-b\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2>a^2-2ab+b^2\)

\(\Leftrightarrow4ab>0\Leftrightarrow ab>0\)

\(\Rightarrow a;b\) cùng dấu

Biến đổi VP

=> VT = VP

=> Đpcm

a: \(\Leftrightarrow\left(a+1\right)^2-4a\ge0\)

hay \(\left(a-1\right)^2>=0\)(luôn đúng)

b: \(VT=a^2c^2+2abcd+b^2d^2+a^2d^2-2abcd+b^2c^2\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(c^2+d^2\right)\left(a^2+b^2\right)=VP\)