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Vì \(ab+bc+ac=3\) => \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{abc}\)
Đặt \(\frac{1}{a}=x\): \(\frac{1}{b}=y\): \(\frac{1}{c}=z\)=> x+y+z=3xyz
Ta có \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{xyz}\ge13\)
AD BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) dấu = khi a=b=c ta có
\(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{36}{x+y+z}\)=\(\frac{36}{3xyz}=\frac{12}{xyz}\)
=> \(\frac{12}{xyz}+\frac{1}{xyz}\ge13\)
=> \(\frac{13}{xyz}\ge13\)
mà \(3xyz=x+y+z\ge3\sqrt[3]{xyz}\)dấu = khi x=y=z
=> xyz\(\le1\)
=> đpcm
Ta có
\(\frac{\left(a+b+c\right)^2}{3}\)> ab + bc + ca =3 => a + b + => 3
ta có abc > ( a+b+c) ( b + c -a ) ( c + a -b)
= ( a+b+c+ 2c) ( b + c -a +2a) ( c + a -b+2b)
> ( 3 -2c ) ( 3 - 2 a ) ( 3 - 2 b ) ( do a+b + c)> 3
= 12 ( xy + yz + zx ) -8 xyz - 18 ( x + y + z ) + 27
= 12 .3 - 8xyz - 18 .3 +27
9 - 8 xyz
ta có : xyz > 9 - 8 xyz + 8 xyz > 9 => xyz > 1
do đó : 4 ( a + b + c ) + abc > 4.3 + 1 = 13 (dpcm)
hok tốt
Ta có
\(\frac{\left(a+b+c\right)^2}{3}\)> ab + bc + ca =3 => a + b + => 3
ta có abc > ( a+b+c) ( b + c -a ) ( c + a -b)
= ( a+b+c+ 2c) ( b + c -a +2a) ( c + a -b+2b)
> ( 3 -2c ) ( 3 - 2 a ) ( 3 - 2 b ) ( do a+b + c)> 3
= 12 ( xy + yz + zx ) -8 xyz - 18 ( x + y + z ) + 27
= 12 .3 - 8xyz - 18 .3 +27
9 - 8 xyz
ta có : xyz > 9 - 8 xyz + 8 xyz > 9 => xyz > 1
do đó : 4 ( a + b + c ) + abc > 4.3 + 1 = 13 (dpcm)
hok tốt
1)
Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)
Dấu "=" xảy ra khi a=b=c
2)
\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)
Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
Ta có
\(\frac{\left(a+b+c\right)^2}{3}\)> ab + bc + ca =3 => a + b + => 3
ta có abc > ( a+b+c) ( b + c -a ) ( c + a -b)
= ( a+b+c+ 2c) ( b + c -a +2a) ( c + a -b+2b)
> ( 3 -2c ) ( 3 - 2 a ) ( 3 - 2 b ) ( do a+b + c)> 3
= 12 ( xy + yz + zx ) -8 xyz - 18 ( x + y + z ) + 27
= 12 .3 - 8xyz - 18 .3 +27
9 - 8 xyz
ta có : xyz > 9 - 8 xyz + 8 xyz > 9 => xyz > 1
do đó : 4 ( a + b + c ) + abc > 4.3 + 1 = 13 (dpcm)
hok tốt