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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 \(\left(4a;5b;-6c\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x+y+z=-5\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y+z\right)^2=25\\\frac{xy+yz+zx}{xyz}=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+z^2+2\left(xy+yz+zx\right)=25\\xy+yz+zx=0\end{matrix}\right.\)
\(\Rightarrow x^2+y^2+z^2=25\) hay \(16a^2+25b^2+36c^2=25\)
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Do a ; b ; c > 0 ( GT )
Áp dụng BĐT phụ \(3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\) , ta có :
\(3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow12\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow3+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le1\)
Lại có : \(\frac{1}{4a+b+c}=\frac{1}{a+a+a+a+b+c}\le\frac{1}{36}\left(\frac{4}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(1\right)\)
( áp dụng BĐT phụ \(\frac{1}{a1}+\frac{1}{a2}+\frac{1}{a3}+\frac{1}{a4}+\frac{1}{a5}+\frac{1}{a6}\ge\frac{36}{a1+a2+a3+a4+a5+a6}\) )
CMTT , ta có : \(\frac{1}{4b+a+c}\le\frac{1}{36}\left(\frac{4}{b}+\frac{1}{a}+\frac{1}{c}\right);\frac{1}{4c+a+b}\le\frac{1}{36}\left(\frac{4}{c}+\frac{1}{a}+\frac{1}{b}\right)\left(2\right)\)
Từ ( 1 ) ; ( 2 ) \(\Rightarrow\frac{1}{4a+b+c}+\frac{1}{4b+a+c}+\frac{1}{4c+a+b}\le\frac{1}{36}\left(\frac{6}{a}+\frac{6}{b}+\frac{6}{c}\right)=\frac{1}{6}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le\frac{1}{6}.1=\frac{1}{6}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=3\)
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a) Ta có:
\(\dfrac{a^2}{a-1}\) \(\geq\) 4(*)
\(\Leftrightarrow\) a2 \(\geq\) 4.(a-1)
\(\Leftrightarrow\) a2 \(\geq\) 4a-4
\(\Leftrightarrow\) a2-4a+4 \(\geq\) 0
\(\Leftrightarrow\) (a-2)2 \(\geq\) 0(**)
Ta có BĐT(**) luôn đúng nên suy ra BĐT(*) luôn đúng
Dấu = xảy ra khi và chỉ khi a=2
B) Áp dụng câu a ta được:
\(\dfrac{4a^2}{a-1}=4.\dfrac{a^2}{a-1}\) \(\geq\) 4.4=16(1)
\(\dfrac{5b^2}{b-1}=5.\dfrac{b^2}{b-1}\) \(\geq\) 5.4=20(2)
\(\dfrac{3c^2}{c-1}=3.\dfrac{c^2}{c-1}\) \(\geq\) 3.4=12(3)
Cộng các BĐT(1),(2),(3) ta được
\(\dfrac{4a^2}{a-1}+\dfrac{5b^2}{b-1}+\dfrac{3c^2}{c-1}\) \(\geq\) 16+20+12=48
Dấu = xảy ra khi và chỉ khi a=b=c=2
Đặt A= \(\dfrac{4a^2}{a-1}+\dfrac{8b^2}{b-1}+\dfrac{12c^2}{c-1}\)
Áp dụng BĐT đã CM ta có:
A= \(\dfrac{4a^2}{a-1}+\dfrac{8b^2}{b-1}+\dfrac{12c^2}{c-1}\) \(\geq\) 4.4+8.4+12.4=16+32+48=96
\(\Rightarrow\) \(\dfrac{4a^2}{a-1}+\dfrac{8b^2}{b-1}+\dfrac{12c^2}{c-1}\) \(\geq\) 96
hay A \(\geq\) 96
Dấu = xảy ra khi và chỉ khi a=b=c=2
Vậy MinA=96 khi và chỉ khi a=b=c=2
a)
Ta có :
\(\dfrac{a^2}{a-1}\ge4\) (1)
\(\Leftrightarrow\dfrac{a^2}{a-1}\ge\dfrac{4a-4}{a-1}\left(\forall a-1\ne0\right)\)
\(\Leftrightarrow a^2\ge4a-4\)
\(\Leftrightarrow a^2-4a+4\ge0\)
\(\Leftrightarrow\left(a-2\right)^2\ge0\)(luôn đúng) (2)
BĐT (2) đúng suy ra BĐT (1) luôn đúng
Dấu bằng xảy ra chỉ khi và khi a = 2
Ta có : ( x - 2 )2 \(\ge\)0 \(\Leftrightarrow\)x2 - 4x + 4 \(\ge\)0
\(\Rightarrow\) x2 \(\ge\)4x - 4 \(\Rightarrow\)x2 \(\ge\)4 . ( x - 1 ) \(\Rightarrow\)\(\frac{x^2}{x-1}\)\(\ge\)4
\(\Rightarrow\frac{4a^2}{a-1}+\frac{5b^2}{b-1}+\frac{3c^2}{c-1}\ge4.4+5.4+3.4=48\)