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Ta có: BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)( CM bằng BĐT Shwars nha).Áp dụng ta có:
\(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5a}+\frac{1}{3a+2b+4c}\ge\frac{9}{9a+6b+12c}=\frac{3}{3a+2b+4c}\left(1\right)\)
\(\frac{1}{b+3c+5a}+\frac{1}{c+3a+5b}+\frac{1}{3b+2c+4a}\ge\frac{9}{9b+6c+12a}=\frac{3}{3b+2c+4a}\left(2\right)\)
\(\frac{1}{c+3a+5b}+\frac{1}{a+3b+5c}+\frac{1}{3c+2a+4b}\ge\frac{9}{9c+6a+12b}=\frac{3}{3c+2a+4b}\left(3\right)\)
Cộng (1),(2) và (3) có:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)
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Đặt \(\hept{\begin{cases}3a+b-c=x\\3b+c-a=y\\3c+a-b=z\end{cases}}\)
Khi đó điều kiện đb tương ứng
\(\left(x+y+z\right)^3=24+x^3+y^3+z^3\)
\(\Leftrightarrow3.\left(x+y\right).\left(x+z\right).\left(x+z\right)=24\)
\(\Rightarrow3.\left(2a+4b\right).\left(2b+4c\right).\left(2c+4a\right)=24\)
\(\Rightarrow\left(a+2b\right).\left(b+2c\right).\left(c+2a\right)=1\)
Do đó ta có đpcm
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Hình như đề sai , giả sử a = b = c = 0
=> vế trái bằng 0 , vé phải bằng 24
\(\left(3a+b-c\right)^3+\left(3b+c-a\right)^3+\left(3c+a-b\right)^3+24\)
\(=24+27a^3+27b^3+27c^3+3\left(\left(3a+b\right)\left(3a-c\right)\left(b-c\right)+\left(3b+c\right)\left(3b-a\right)\left(c-a\right)+\left(3c+a\right)\left(3c-b\right)\left(a-b\right)\right)\)\(\left(3a+3b+3c\right)^3=27a^3+27b^3+27c^3+81\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow8+A=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
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Lời giải:
Đặt ⎧⎪⎨⎪⎩3a+b−c=x3b+c−a=y3c+a−b=z{3a+b−c=x3b+c−a=y3c+a−b=z
Khi đó, điều kiện đb tương đương với:
(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24
⇔3(2a+4b)(2b+4c)(2c+4a)=24⇔3(2a+4b)(2b+4c)(2c+4a)=24
⇔(a+2b)(b+2c)(c+2a)=1⇔(a+2b)(b+2c)(c+2a)=1
Do đó ta có đpcm
Lời giải:
Đặt ⎧⎪⎨⎪⎩3a+b−c=x3b+c−a=y3c+a−b=z{3a+b−c=x3b+c−a=y3c+a−b=z
Khi đó, điều kiện đb tương đương với:
(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24
⇔3(2a+4b)(2b+4c)(2c+4a)=24⇔3(2a+4b)(2b+4c)(2c+4a)=24
⇔(a+2b)(b+2c)(c+2a)=1⇔(a+2b)(b+2c)(c+2a)=1
Do đó ta có đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}< =\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\)
Có \(\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}< =\dfrac{4}{2a+4b+2c}=\dfrac{2}{a+2b+c}\)
Cm tương tự, ta có:
\(\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}< =\dfrac{2}{b+2c+a}\)\(\)
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}< =\dfrac{2}{c+2a+b}\)
Cộng 2 vế của 3 BĐT với nhau, ta có:
\(\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}+\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}+\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}< =\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\)
\(\Leftrightarrow\left(\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}+\dfrac{1}{a+2b+c}\right)+\left(\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\right)< =\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\)
\(\Leftrightarrow\dfrac{-\left(c+2a+b\right)\cdot\left(a+2b+c\right)-\left(b+2c+a\right)\left(a+2b+c\right)-\left(b+2c+a\right)\left(c+2a+b\right)}{\left(b+2c+a\right)\cdot\left(c+2a+b\right)\cdot\left(a+2b+c\right)}+\dfrac{\left(b+3c\right)\left(c+3a\right)+\left(a+3b\right)\left(c+3a\right)+\left(a+3b\right)\left(b+3c\right)}{\left(a+3b\right)\left(b+3c\right)\left(c+3a\right)}\le0\)
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Áp dụng bất đẳng thức Cauchy-Schwarz:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{c+3a}\ge\dfrac{\left(1+1\right)^2}{a+2b+c+c+3a}=\dfrac{4}{4a+2b+2c}=\dfrac{2}{2a+b+c}\)
Chứng minh tương tự ta được: \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+a}+\dfrac{1}{a+3b}\ge\dfrac{2}{a+2b+c}\\\dfrac{1}{c+2a+b}+\dfrac{1}{b+3c}\ge\dfrac{2}{a+b+2c}\end{matrix}\right.\)
Cộng theo vế:
\(\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}+\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{2}{a+2b+c}+\dfrac{2}{b+2c+a}+\dfrac{2}{c+2a+b}\)
\(\Rightarrow\dfrac{1}{a+3b}+\dfrac{1}{b+3c}+\dfrac{1}{c+3a}\ge\dfrac{1}{a+2b+c}+\dfrac{1}{b+2c+a}+\dfrac{1}{c+2a+b}\)
p/s: đã sửa đề