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Áp dụng bất đẳng thức \(\dfrac{9}{x+y+z}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) với x, y, z > 0 ta có:
\(\dfrac{1}{2a+b}+\dfrac{1}{2b+c}+\dfrac{1}{2c+a}=\dfrac{1}{9}\left(\dfrac{9}{a+a+b}+\dfrac{9}{b+b+c}+\dfrac{1}{c+c+a}\right)\le\dfrac{1}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{c}+\dfrac{1}{a}\right)=\dfrac{1}{9}.3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}\).
Theo BĐT Bu nhi a cốp xki ta có :
\(\left(a+b+c+d\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\ge16\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{16}{a+b+c+d}\)
Áp dụng vào bài toán ta có :
\(\dfrac{1}{3a+3b+2c}=\dfrac{1}{16}.\dfrac{16}{\left(a+b\right)+\left(a+b\right)+\left(b+c\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3b+3c+2a}=\dfrac{1}{16}.\dfrac{16}{\left(b+c\right)+\left(b+c\right)+\left(a+b\right)+\left(c+a\right)}\le\dfrac{1}{16}\left(\dfrac{1}{b+c}+\dfrac{1}{b+c}+\dfrac{1}{a+b}+\dfrac{1}{c+a}\right)\)
\(\dfrac{1}{3c+3a+2b}=\dfrac{1}{16}.\dfrac{16}{\left(c+a\right)+\left(c+a\right)+\left(a+b\right)+\left(b+c\right)}\le\dfrac{1}{16}\left(\dfrac{1}{c+a}+\dfrac{1}{c+a}+\dfrac{1}{a+b}+\dfrac{1}{b+c}\right)\)
Cộng từng vế của BĐT ta được :
\(\dfrac{1}{3a+3b+2c}+\dfrac{1}{3b+3c+2a}+\dfrac{1}{3c+3a+2b}\le\dfrac{1}{16}\left(\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\right)=\dfrac{1}{4}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=\dfrac{1}{4}.6=\dfrac{3}{2}\)
Vậy GTLN của A là \(\dfrac{3}{2}\) . Dấu \("="\) xảy ra khi \(a=b=c=\dfrac{1}{4}\)
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:
\(2\left(\frac{1}{a+3b+5c}+\frac{1}{b+3c+5c}+\frac{1}{c+3a+5b}\right)+\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\ge3\left(\frac{1}{3a+2b+4c}+\frac{1}{3b+2c+4a}+\frac{1}{3c+2a+4b}\right)\)
\(\Rightarrow2VP\ge2VT\)
\(\RightarrowĐPCM\)
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
Ta có:\(\dfrac{1}{1+ab}+\dfrac{1}{1+bc}+\dfrac{1}{1+ac}\ge\dfrac{9}{1+1+1+ab+bc+ca}\)(AM-GM)
Lại có:\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow\dfrac{9}{3+ab+bc+ca}\ge\dfrac{9}{3+a^2+b^2+c^2}=\dfrac{9}{6}=\dfrac{3}{2}\)
\(\Rightarrowđpcm\)
Cháu làm cho bác câu 2 thôi,câu 3 THANGDZ làm rồi sợ mất bản quyền lắm:v
Lời giải:
Áp dụng liên tiếp bất đẳng thức AM-GM và Cauchy-Schwarz ta có:
\(\dfrac{a}{a+2b+3c}+\dfrac{b}{b+2c+3a}+\dfrac{c}{c+2a+3b}\)
\(=\dfrac{a^2}{a^2+2ab+3ac}+\dfrac{b^2}{b^2+2bc+3ab}+\dfrac{c^2}{c^2+2ac+3bc}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+5ab+5bc+5ac}\)
\(=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+3\left(ab+bc+ac\right)}\ge\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)^2+\left(a+b+c\right)^2}=\dfrac{1}{2}\)
Ta có:
A = \(\frac{a}{2b+3c}+\frac{b}{2c+3a}+\frac{c}{3b+2a}=\frac{a^2}{2ab+3ac}+\frac{b^2}{2bc+3ab}+\frac{c^2}{3bc+2ac}\)
A \(\ge\frac{\left(a+b+c\right)^2}{2ab+3ac+2bc+3ab+3bc+2ac}\)(bđt svacxo \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\))
A \(\ge\frac{\left(a+b+c\right)^2}{5\left(ab+bc+ac\right)}\ge\frac{\left(a+b+c\right)^2}{\frac{5\left(a+b+c\right)^2}{3}}\) (bđt \(xy+yz+xz\le\frac{\left(x+y+z\right)^2}{3}\)(*)
CM bđt * <=> \(3xy+3yz+3xz\le x^2+y^2+z^2+2xz+2xy+2yz\)
<=> \(\left(x-y\right)^2+\left(x-z\right)^2+\left(y-z\right)^2\ge0\) (luôn đúng)
<=> A \(\ge\frac{3}{5}\) --> ĐPCM
