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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
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
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: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)
\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)
\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)
\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)
\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)
\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)
\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)
Vậy VT = VP, đẳng thức được chứng minh
Ta có: \(abc=1\Leftrightarrow\hept{\begin{cases}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ca=\frac{1}{b}\end{cases}}\)
\(abc=1\Leftrightarrow\sqrt[3]{abc}=1\)
Áp dụng BĐT AM-GM ta có:\(1=\sqrt[3]{abc}\le\frac{a+b+c}{3}\Leftrightarrow a+b+c\ge3\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge4\left(a+b+c-1\right)\)
\(\Leftrightarrow\)\(a^2b+ab^2+a^2c+ac^2+b^2c+cb^2+2abc+4\ge4\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a}{c}+\frac{b}{c}+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{c}{a}+6\ge4\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}+6\ge4\left(a+b+c\right)\)
\(\Leftrightarrow\frac{a+b+c}{c}+\frac{a+c+b}{b}+\frac{a+b+c}{a}+3\ge4\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge4\left(a+b+c\right)\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\ge4\)(1)
Ta chứng mĩnh BĐT phụ
Với a,b,c > thì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)
Thật vậy.
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)
Áp dụng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\ge\frac{9}{a+b+c}+\frac{3}{a+b+c}=\frac{12}{3}=4\)(2)
Từ (1) và (2)
=> \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge4\left(a+b+c-1\right)\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)
Bạn ơi, tại sao \(\frac{9}{a+b+c}+\frac{3}{a+b+c}=\frac{12}{3}\) được hả bạn?
Lời giải:
Vì $abc=1$ nên:
\((a+bc)(b+ac)(c+ab)=a(a+bc)b(b+ac)c(c+ab)=(a^2+1)(b^2+1)(c^2+1)\)
Áp dụng BĐT Bunhiacopxky:
\((a^2+1)(1+b^2)\geq (a+b)^2; (a^2+1)(1+c^2)\geq (a+c)^2; (b^2+1)(1+c^2)\geq (b+c)^2\)
Nhân theo vế và thu gọn:
\(\Rightarrow (a^2+1)(b^2+1)(c^2+1)\geq (a+b)(b+c)(c+a)\)
Lại có: Theo BĐT AM-GM thì:
\((a+b)(b+c)(c+a)=(ab+bc+ac)(a+b+c)-abc\)
\(\geq (ab+bc+ac)(a+b+c)-\frac{(a+b+c)(ab+bc+ac)}{9}=\frac{8(a+b+c)(ab+bc+ac)}{9}(*)\) (đây là BĐT khá quen thuộc rồi)
Do đó:
\(P=\frac{(a+bc)(b+ca)(c+ab)}{ab+bc+ac}+\frac{1}{a+b+c}=\frac{(a^2+1)(b^2+1)(c^2+1)}{ab+bc+ac}+\frac{1}{a+b+c}\geq \frac{(a+b)(b+c)(c+a)}{ab+bc+ac}+\frac{1}{a+b+c}\)
\(P\geq \frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\)
Áp dụng BĐT (*) và AM-GM:
\(\frac{7(a+b)(b+c)(c+a)}{8(ab+bc+ac)}\geq 7.\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(ab+bc+ac)}=\frac{7}{9}(a+b+c)\geq \frac{7}{9}.3\sqrt[3]{abc}=\frac{7}{3}\)
\(\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)}+\frac{1}{a+b+c}\geq 2\sqrt{\frac{(a+b)(b+c)(c+a)}{8(ab+bc+ac)(a+b+c)}}\geq 2\sqrt{\frac{\frac{8}{9}(a+b+c)(ab+bc+ac)}{8(a+b+c)(ab+bc+ac)}}=\frac{2}{3}\)
\(\Rightarrow P\geq \frac{7}{3}+\frac{2}{3}=3\)
Vậy $P_{\min}=3$
\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\)
\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1\)
\(=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2+1+1+1-1\)
Áp dụng BĐT AM-GM ta có:
\(\left(a+bc\right)\left(b+ca\right)\left(c+ab\right)\ge a^2+b^2+c^2+2ab+2bc+2ac-1=\left(a+b+c\right)^2-1\)\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\)
Dấu " = " xảy ra <=> ...
Ta có: \(\frac{1}{3}.\left(a+b+c\right)^2\ge ab+bc+ca\)( BĐT quen thuộc tự c/m)
\(\Rightarrow P\ge\frac{\left(a+b+c\right)^2-1}{ab+bc+ca}+\frac{1}{a+b+c}\ge\frac{\left(a+b+c\right)^2}{\frac{1}{3}\left(a+b+c\right)^2}-\frac{1}{\frac{1}{3}\left(a+b+c\right)}+\frac{1}{a+b+c}\)\(=3+\frac{a+b+c-3}{\left(a+b+c\right)^2}\)
Ta có: \(abc=1\Leftrightarrow\sqrt[3]{abc}=1\le\frac{a+b+c}{3}\left(AM-GM\right)\)
\(\Rightarrow a+b+c\ge3\)
Dấu " = " xảy ra <=> ...
\(\Rightarrow P\ge3+\frac{a+b+c-3}{\left(a+b+c\right)^2}\ge3\)
Dấu " = " xảy ra <=> a=b=c=1
KL:...........
Ta có:\(3\left(\frac{ab+bc+ca}{a+b+c}\right)^2\le3\left[\frac{\frac{\left(a+b+c\right)^2}{3}}{a+b+c}\right]^2\)\(=3\left(\frac{a+b+c}{3}\right)^2=\frac{\left(a+b+c\right)^2}{3}\le a^2+b^2+c^2\)(1)
Mặt khác:\(\left(\frac{ab}{c}\right)^2+\left(\frac{bc}{a}\right)^2\ge2.\frac{ab}{c}.\frac{bc}{a}=2b^2\)(2)
Tương tự ta cũng có:\(\left(\frac{bc}{a}\right)^2+\left(\frac{ca}{b}\right)^2\ge2c^2\)(3);\(\left(\frac{ca}{b}\right)^2+\left(\frac{ab}{c}\right)^2\ge2a^2\)(4)
Cộng theo vế (1),(2),(3) ta được:\(2\left[\left(\frac{ab}{c}\right)^2+\left(\frac{bc}{a}\right)^2+\left(\frac{ca}{b}\right)^2\right]\ge2\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\left(\frac{ab}{c}\right)^2+\left(\frac{bc}{a}\right)^2+\left(\frac{ca}{b}\right)^2\ge a^2+b^2+c^2\)(5)
Từ (1) và (5) suy ra điều phải chứng minh.Dấu "=" xảy ra khi \(a=b=c\)
cảm ơn. Nghĩ hộ mình nhé!
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