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Áp dụng BĐT:1/a+1/b>=4/a+b
Ta có:
1/(p-a)+1/(p+b)>=4/(2p-a-b)=4/c
Các phần sau tương tự!
=>2VT>=4(1/a+1/b+1/c)
=>VT>=2(1/a+1/b+1/c)
b)
Dấu "=" xảy ra p-a=p-b=p-c => a=b=c
=>tg đều
Áp dungj BĐt Cauchy - Schwarz :
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\)
\(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{2p-b-c}=\frac{4}{a}\)
\(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{2p-b-c}=\frac{4}{a}\)
Cộng theo vế và thu gọn ta được :
\(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Ta có : đpcm
Dấu " = " xảy ra khi \(a=b=c\)
Ta có
\(P=\frac{a+b+c}{2}\Rightarrow2p=a+b+c\)
áp dụng bđt Cauchy-Schwarz ta có
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{p-a+p-b}=\frac{4}{2p-a-b}=\frac{4}{a+b+c-a-b}=\frac{4}{c}\left(1\right)\)
C/m tương tự ta có
\(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\left(2\right)\)
\(\frac{1}{p-a}+\frac{1}{p-c}\ge\frac{4}{b}\left(3\right)\)
Cộng vế theo vế (1) (2) và (3) => đpcm
Dễ dàng CM BĐT sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b},\forall a,b>0\)
Áp dung: \(\hept{\begin{cases}\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-a-b}=\frac{4}{c}\\\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{2p-b-c}=\frac{4}{a}\\\frac{1}{p-c}+\frac{1}{p-a}\ge\frac{4}{2p-c-a}=\frac{4}{b}\end{cases}}\)
Cộng vế theo vế các BĐT trên => ĐPCM
Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)
\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)
Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Ai có cách hay?
1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.
2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)
\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)
\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)
4. Ta có: \(a+b+c=6abc\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=6\)
Đặt \(\frac{1}{a}=x;\frac{1}{b}=y;\frac{1}{c}=z\)
\(\Rightarrow xy+yz+zx=6\)
Lại có: \(\frac{bc}{a^3\left(c+2b\right)}=\frac{1}{a^3\frac{c+2b}{bc}}=\frac{\frac{1}{a^3}}{\frac{1}{b}+\frac{2}{c}}=\frac{x^3}{y+2z}\)
Tương tự suy ra:
\(S=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2zx}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{x^2+y^2+z^2}{3}\ge\frac{xy+yz+zx}{3}=2\)
Dấu = xảy ra khi \(x=y=z=\sqrt{2}\Rightarrow a=b=c=\frac{1}{\sqrt{2}}\)
\(\hept{\begin{cases}\frac{ab}{c}+\frac{bc}{a}\ge2b\\\frac{bc}{a}+\frac{ca}{b}\ge2c\\\frac{ca}{b}+\frac{ab}{c}\ge2a\end{cases}}\) :)))
a^2+b^2+c^2=(a+b+c)^2-2ab-2bc-2ca=1-2ab-2bc-2ca
((a^2+b^2+c^2)-1)/2abc=(1-2ab-2bc-2ca-1)/abc=-(1/a+1/b+1/c)
T=4/a+b +4/b+c +4/c+a<=1/a+1/b+1/b+1/c+1/c+1/a-1/a-1/b-1/c=1/a+1/b+1/c<=9
Dấu = khi a=b=c=1/3
\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\Leftrightarrow\frac{a+b}{ab}\ge\frac{4}{a+b}\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\Leftrightarrow\left(a-b\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b\)
b/ Áp dụng BĐT ở câu a:
\(\frac{1}{p-a}+\frac{1}{p-b}\ge\frac{4}{2p-\left(a+b\right)}=\frac{4}{c}\)
Tương tự: \(\frac{1}{p-b}+\frac{1}{p-c}\ge\frac{4}{a}\) ; \(\frac{1}{p-a}+\frac{1}{p-c}\ge\frac{4}{b}\)
Cộng vế với vế: \(2\left(\frac{1}{p-a}+\frac{1}{p-b}+\frac{1}{p-c}\right)\ge2\left(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c\)
c/ \(2p=a+b+c=18\)
\(\Rightarrow a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2=\frac{18^2}{3}=108\)
Dấu "=" xảy ra khi \(a=b=c=6\)