Chứng minh rằng với mọi số thực a;b;c thì
|a|+|b|+|c|+|a+b|+|b+c|+|c+a| ≥ 1/3.(|5a+4b|+|5b+4c|+|5c+4a|)
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DV
1
2 tháng 11 2021
\(\left(a+2\right)^2+\left(a+4\right)^2=a^2+4a+4+a^2+8a+16\)
\(=2a^2+12a+20=2\left(a^2+6a+9\right)+2=2\left(a+3\right)^2+2\ge2>0\forall a\in R\)
25 tháng 8 2021
Điều cần chứng minh:
|a|+|b|≥|a+b||a|+|b|≥|a+b|
|a+b|=|a+b||a+b|=|a+b|
Khi này ,a và b có thể nhận với giá trị âm hoặc dương hoặc bằng 0
|a|>=0. và |b|>=0
Nên chúng chỉ có nhận giá trị lớn hơn or bằng 0
⇒|a|+|b|≥|a+b|→đpcm
25 tháng 8 2021
\(\left\{{}\begin{matrix}\left|a\right|>=0\\\left|b\right|>=0\end{matrix}\right.\)
NT
0
NT
0
NT
0
2 tháng 11 2021
\(\left(a+2\right)^2+\left(b+2\right)^2+\left(a^2+b^2+ab\right)\\ =a^2+4a+4+b^2+4b+4+a^2+b^2+ab\\ =2a^2+2b^2+4a+4b+ab+8\\ =\left[\left(a^2+ab+\dfrac{1}{4}b^2\right)+2\left(a+\dfrac{1}{2}b\right)+1\right]+\left(a^2+2a+1\right)+\dfrac{7}{4}\left(b^2+2\cdot\dfrac{6}{7}b+\dfrac{42}{49}\right)+\dfrac{9}{2}\\ =\left(a+\dfrac{1}{2}b+1\right)^2+\left(a+1\right)^2+\dfrac{7}{4}\left(b+\dfrac{6}{7}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}>0\left(đpcm\right)\)
20 tháng 3 2018
a)\(a^2+ab+b^2=a^2+\dfrac{2ab}{2}+\left(\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\)
\(=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\forall a,b\)
b)\(a^4+b^4\ge a^3b+ab^3\)
\(\Leftrightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a^3-b^3\right)\left(a-b\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\forall a,b\)
11 tháng 12 2019
a) Đề sai thì phải.Phải là CM: \(x^2-x+1>0\) với mọi x
Ta có:
\(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\) nên \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
Vậy \(x^2-x+1>0\) với mọi \(x\in R\)
b)Ta có:
\(-x^2+2x-4=-\left(x^2-2x+1\right)-3\)
\(=-\left(x-1\right)^2-3\)
Vì \(-\left(x-1\right)^2\le0\) với mọi x nên \(-\left(x-1\right)^2-3< 0\)
Vậy \(-x^2+2x-4< 0\) với mọi \(x\in R\)
- Bất đẳng thức tam giác: |x|+|y|≥|x+y|the absolute value of x end-absolute-value plus the absolute value of y end-absolute-value is greater than or equal to the absolute value of x plus y end-absolute-value|𝑥|+|𝑦|≥|𝑥+𝑦|.
- Bất đẳng thức tam giác mở rộng: |x|+|y|+|z|≥|x+y+z|the absolute value of x end-absolute-value plus the absolute value of y end-absolute-value plus the absolute value of z end-absolute-value is greater than or equal to the absolute value of x plus y plus z end-absolute-value|𝑥|+|𝑦|+|𝑧|≥|𝑥+𝑦+𝑧|.
Cách giải Sử dụng bất đẳng thức tam giác để chứng minh từng phần của bất đẳng thức.- Bước 1 . Chứng minh |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13|5a+4b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction the absolute value of 5 a plus 4 b end-absolute-value|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13|5𝑎+4𝑏|.
- Ta có 3(|a|+|b|+|a+b|)≥|3a+3b|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b end-absolute-value3(|𝑎|+|𝑏|+|𝑎+𝑏|)≥|3𝑎+3𝑏|.
- Ta có 3(|a|+|b|+|c|+|a+b|+|b+c|+|c+a|)≥|3a+3b+3c+3a+3b+3c|=|6a+6b+6c|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b plus 3 c plus 3 a plus 3 b plus 3 c end-absolute-value equals the absolute value of 6 a plus 6 b plus 6 c end-absolute-value3(|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|)≥|3𝑎+3𝑏+3𝑐+3𝑎+3𝑏+3𝑐|=|6𝑎+6𝑏+6𝑐|.
- Áp dụng bất đẳng thức tam giác:
- |a|+|a|+|a|+|b|+|b|≥|3a+2b|the absolute value of a end-absolute-value plus the absolute value of a end-absolute-value plus the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of b end-absolute-value is greater than or equal to the absolute value of 3 a plus 2 b end-absolute-value|𝑎|+|𝑎|+|𝑎|+|𝑏|+|𝑏|≥|3𝑎+2𝑏|.
- |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- Xét |5a+4b|=|(a+b)+(a+b)+(a+b)+a+a+b|the absolute value of 5 a plus 4 b end-absolute-value equals the absolute value of open paren a plus b close paren plus open paren a plus b close paren plus open paren a plus b close paren plus a plus a plus b end-absolute-value|5𝑎+4𝑏|=|(𝑎+𝑏)+(𝑎+𝑏)+(𝑎+𝑏)+𝑎+𝑎+𝑏|.
- Ta có |5a+4b|=|(a+b)+(a+b)+(a+b)+a+a+b|the absolute value of 5 a plus 4 b end-absolute-value equals the absolute value of open paren a plus b close paren plus open paren a plus b close paren plus open paren a plus b close paren plus a plus a plus b end-absolute-value|5𝑎+4𝑏|=|(𝑎+𝑏)+(𝑎+𝑏)+(𝑎+𝑏)+𝑎+𝑎+𝑏|.
- Áp dụng bất đẳng thức tam giác:
- |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13|5a+4b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction the absolute value of 5 a plus 4 b end-absolute-value|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13|5𝑎+4𝑏|.
- Ta có 3(|a|+|b|+|a+b|)≥|3a+3b|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value close paren is greater than or equal to the absolute value of 3 a plus 3 b end-absolute-value3(|𝑎|+|𝑏|+|𝑎+𝑏|)≥|3𝑎+3𝑏|.
- Ta có 3(|a|+|b|+|c|+|a+b|+|b+c|+|c+a|)≥|5a+4b|+|5b+4c|+|5c+4a|3 open paren the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value close paren is greater than or equal to the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value3(|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|)≥|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|.
- Ta có |a|+|b|+|a+b|≥|2a+2b|the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of a plus b end-absolute-value is greater than or equal to the absolute value of 2 a plus 2 b end-absolute-value|𝑎|+|𝑏|+|𝑎+𝑏|≥|2𝑎+2𝑏|.
- Ta có |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13(|5a+4b|+|5b+4c|+|5c+4a|)the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction open paren the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value close paren|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13(|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|).
Lời giải Bất đẳng thức cần chứng minh là |a|+|b|+|c|+|a+b|+|b+c|+|c+a|≥13(|5a+4b|+|5b+4c|+|5c+4a|)the absolute value of a end-absolute-value plus the absolute value of b end-absolute-value plus the absolute value of c end-absolute-value plus the absolute value of a plus b end-absolute-value plus the absolute value of b plus c end-absolute-value plus the absolute value of c plus a end-absolute-value is greater than or equal to 1 over 3 end-fraction open paren the absolute value of 5 a plus 4 b end-absolute-value plus the absolute value of 5 b plus 4 c end-absolute-value plus the absolute value of 5 c plus 4 a end-absolute-value close paren|𝑎|+|𝑏|+|𝑐|+|𝑎+𝑏|+|𝑏+𝑐|+|𝑐+𝑎|≥13(|5𝑎+4𝑏|+|5𝑏+4𝑐|+|5𝑐+4𝑎|).