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a) Từ giả thiết : \(a^2+2c^2=3b^2+19\Rightarrow a^2+2c^2-3b^2=19\)
Ta có : \(\frac{a^2+7}{4}=\frac{b^2+6}{5}=\frac{c^2+3}{6}=\frac{3b^2+18}{15}=\frac{2c^2+6}{12}\)\(=\frac{a^2+7+2c^2+6-3b^2-18}{4+12-15}=\frac{14}{1}=14\)
\(\Rightarrow\)\(a^2=49\Rightarrow a=7\)
\(\Rightarrow\)\(b^2=64\Rightarrow b=8\)
\(\Rightarrow\)\(c^2=81\Rightarrow c=9\)
b) \(P=x^4+2x^3+3x^2+2x+1\)
\(=\left(x^4+2x^2+1\right)+\left(2x^3+2x\right)+x^2=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)
\(=\left(x^2+x+1\right)^2\)
Vì \(x^2+x+1=\left(x^2+2x\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Nên \(P\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)
Dấu bằng xảy ra khi và chỉ khi \(x=-\frac{1}{2}\)
Ta có : \(p=\frac{bc}{a^2\left(b+c\right)}+\frac{ca}{b^2\left(a+c\right)}+\frac{ab}{c^2\left(a+b\right)}\)
Áp dụng bất đẳng thức AM - GM ta có :
\(\frac{bc}{a^2\left(b+c\right)}+\frac{b+c}{4bc}\ge2\sqrt{\frac{bc}{a^2\left(b+c\right)}.\frac{b+c}{4ab}}=\frac{1}{a}\)
\(\frac{ac}{b^2\left(a+c\right)}+\frac{a+c}{4ac}\ge4\sqrt{\frac{ac}{b^2\left(a+c\right)}.\frac{a+c}{4ac}}=\frac{1}{b}\)
\(\frac{ab}{c^2\left(a+b\right)}+\frac{a+b}{4ab}\ge2\sqrt{\frac{ab}{c^2\left(a+b\right)}.\frac{a+b}{4ab}}=\frac{1}{c}\)
Cộng vế với vế ta được \(p+\frac{1}{4c}+\frac{1}{4a}+\frac{1}{4b}+\frac{1}{4a}+\frac{1}{4c}+\frac{1}{4b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Leftrightarrow p+\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
\(\Rightarrow p\ge\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}\ge3\sqrt[3]{\frac{1}{2a.2b.2c}}=\frac{3}{\sqrt[3]{8abc}}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Xét: \(\frac{bc}{a^2b+ca^2}=\frac{bc}{a\cdot abc\cdot\frac{1}{c}+a\cdot abc\cdot\frac{1}{b}}=\frac{b^2c^2}{ab+ca}\)(*)
Tương tự với (*) ta có: \(\hept{\begin{cases}\frac{ca}{b^2c+ab^2}=\frac{c^2a^2}{ab+bc}\\\frac{ab}{c^2a+bc^2}=\frac{a^2b^2}{ca+bc}\end{cases}}\)
\(\Rightarrow\Sigma_{cyc}\frac{bc}{a^2b+ca^2}=\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\)
Ta thấy\(\Sigma_{cyc}\frac{b^2c^2}{ab+ca}\) có dạng: \(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{1}{2}\left(a+b+c\right)\)
Bước cuối Cô-si ba số và kết hợp điều kiện abc=1 là xong
\(P=\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}=\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}\)
vì a,b,c là 3 cạnh của 1 tam giác áp dụng bđt tam giác có:
\(\hept{\begin{cases}b+c>a\Rightarrow2b+2c>a\Rightarrow2ab+2ac>a^2\Rightarrow2ab+2ac-a^2>0\\c+a>b\Rightarrow2c+2a>b\Rightarrow2bc+2ab>b^2\Rightarrow2bc+2ab-b^2>0\\a+b>c\Rightarrow2a+2b>c\Rightarrow2ac+2bc>c^2\Rightarrow2ac+2bc-c^2>0\end{cases}}\)
\(\Rightarrow\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>0\)áp dụng bđt cauchy schawazt dạng enge ta có:
\(\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>=\)
\(\frac{\left(a+b+c\right)^2}{2ab+2ac-a^2+2bc+2ab-b^2+2ac+2bc-c^2}=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(a^2+b^2+c^2\right)}\left(1\right)\)
vì \(a^2+b^2+c^2>=ab+ac+bc\Rightarrow4ab+4ac+4bc-\left(a^2+b^2+c^2\right)< =\)
\(4ab+4ac+4bc-\left(ab+ac+bc\right)\)mà \(\left(a+b+c\right)^2>0\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(a^2+b^2+c^2\right)}>=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-\left(ab+ac+bc\right)}\)(2)
\(=\frac{\left(a+b+c\right)^2}{4ab+4ac+4bc-ab-ac-bc}=\frac{\left(a+b+c\right)^2}{3ab+3ac+3bc}=\frac{a^2+b^2+c^2+2ab+2ac+2bc}{3ab+3ac+3bc}\)
\(>=\frac{ab+ac+bc+2ab+2ac+2bc}{3ab+3ac+3bc}=\frac{3ab+3ac+3bc}{3ab+3ac+3bc}=1\)(3)
từ (1)(2)(3)\(\Rightarrow\frac{a^2}{2ab+2ac-a^2}+\frac{b^2}{2bc+2ab-b^2}+\frac{c^2}{2ac+2bc-c^2}>=1\)
\(\Rightarrow P=\frac{a}{2b+2c-a}+\frac{b}{2c+2a-b}+\frac{c}{2a+2b-c}>=1\)
dấu = xảy ra khi a=b=c
vậy min P là 1 khi a=b=c
1. Ta có: \(ab+bc+ca=3abc\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
Đặt \(\hept{\begin{cases}\frac{1}{a}=m\\\frac{1}{b}=n\\\frac{1}{c}=p\end{cases}}\) khi đó \(\hept{\begin{cases}m+n+p=3\\M=2\left(m^2+n^2+p^2\right)+mnp\end{cases}}\)
Áp dụng Cauchy ta được:
\(\left(m+n-p\right)\left(m-n+p\right)\le\left(\frac{m+n-p+m-n+p}{2}\right)^2=m^2\)
\(\left(n+p-m\right)\left(n+m-p\right)\le n^2\)
\(\left(p-n+m\right)\left(p-m+n\right)\le p^2\)
\(\Rightarrow\left(m+n-p\right)\left(n+p-m\right)\left(p+m-n\right)\le mnp\)
\(\Leftrightarrow m^3+n^3+p^3+3mnp\ge m^2n+mn^2+n^2p+np^2+p^2m+pm^2\)
\(\Leftrightarrow\left(m+n+p\right)\left(m^2+n^2+p^2-mn-np-pm\right)+6mnp\ge mn\left(m-n\right)+np\left(n-p\right)+pm\left(p-m\right)\)
\(=mn\left(3-p\right)+np\left(3-m\right)+pm\left(3-n\right)\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)-3\left(mn+np+pm\right)+6mnp\ge3\left(mn+np+pm\right)-3mnp\)
\(\Leftrightarrow3\left(m^2+n^2+p^2\right)+9mnp\ge6\left(mn+np+pm\right)\)
\(\Leftrightarrow xyz\ge\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(\Rightarrow M\ge2\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)-\frac{1}{3}\left(m^2+n^2+p^2\right)\)
\(=\frac{5}{3}\left(m^2+n^2+p^2\right)+\frac{2}{3}\left(mn+np+pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m^2+n^2+p^2+2mn+2np+2pm\right)\)
\(=\frac{4}{3}\left(m^2+n^2+p^2\right)+\frac{1}{3}\left(m+n+p\right)^2\)
\(\ge\frac{4}{3}\cdot3+\frac{1}{3}\cdot3^2=4+3=7\)
Dấu "=" xảy ra khi: \(m=n=p=1\Leftrightarrow a=b=c=1\)
1a
\(A=\frac{3}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^4+b^4}{2}\ge\frac{6}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^2+b^2\right)^2}{2}}{2}\)
\(\ge10+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{4}=10+\frac{1}{16}=\frac{161}{16}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(A_{min}=\frac{161}{16}\)
1b.\(B=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+b^2}+\frac{a^8+b^8}{4}\ge\frac{2}{\left(a+b\right)^2}+\frac{4}{\left(a+b\right)^2}+\frac{\frac{\left(a^4+b^4\right)^2}{2}}{4}\)
\(\ge6+\frac{\left[\frac{\left(a^2+b^2\right)^2}{2}\right]^2}{8}\ge6+\frac{\left[\frac{\left(a+b\right)^2}{2}\right]^2}{32}=6+\frac{1}{128}=\frac{769}{128}\)
Dau '=' xay ra khi \(a=b=\frac{1}{2}\)
Vay \(B_{min}=\frac{769}{128}\)khi \(a=b=\frac{1}{2}\)
a) Chứng minh được BĐT \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\)(*)
Dấu "=" xảy ra <=> a=b
Áp dụng BĐT (*) vào bài toán ta có:
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+y}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\\\frac{1}{x+y+2z}=\frac{1}{x+y+z+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Tiếp tục áp dụng BĐT (*) ta có:
\(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right);\frac{1}{y+z}\le\frac{1}{4}\left(\frac{1}{y}+\frac{1}{z}\right);\frac{1}{z+x}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\cdot2\cdot\frac{1}{4}\cdot2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=1\)
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{3}{4}\)
b) áp dụng bđt \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)ta có:
\(\hept{\begin{cases}\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{4}{a+b-c+b+c-a}=\frac{4}{2b}=\frac{2}{b}\\\frac{1}{b+c-a}+\frac{1}{a+c-b}\ge\frac{4}{b+c-a+a+c-b}=\frac{4}{2c}=\frac{2}{c}\\\frac{1}{a+b-c}+\frac{1}{a+c-b}\ge\frac{4}{a+b-c+a+c-b}=\frac{4}{2a}=\frac{2}{a}\end{cases}}\)
Cộng theo vế 3 BĐT ta có:
\(2VT\ge\frac{2}{a}+\frac{2}{b}+\frac{2}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2VP\)
\(\Rightarrow VT\ge VP\)
Đẳng thức xảy ra <=> a=b=c
Lời giải :
\(P=\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\)
\(P=\frac{1}{9}\cdot\left(\frac{9}{a+b+b}+\frac{9}{b+c+c}+\frac{9}{c+a+a}\right)\)
Áp dụng bđt Cauchy dạng \(\frac{9}{x+y+z}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)ta có :
\(P\le\frac{1}{9}\left(\frac{1}{a}+\frac{2}{b}+\frac{1}{b}+\frac{2}{c}+\frac{1}{c}+\frac{2}{a}\right)\)
\(=\frac{1}{9}\left(\frac{3}{a}+\frac{3}{b}+\frac{3}{c}\right)\)
\(=\frac{1}{3}\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
\(=\frac{1}{3}\cdot9=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
Theo Cauchy: \(\frac{1}{a+2b}=\frac{1}{a+b+b}\le\frac{1}{9}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(P\le\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=3\)
Đẳng thức xảy ra khi a = b = c = 1.
Vậy..
làm lại dong cuối:\(A\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{a}\)
Mà:\(2c+b=abc\Rightarrow a=\frac{2c+b}{cb}=\frac{2}{b}+\frac{1}{c}\)
\(\Rightarrow2a=\frac{4}{b}+\frac{2}{c}\)
\(\Rightarrow A\ge2a+\frac{6}{a}\)
Ta có:\(A=\left(\frac{1}{b+c-a}+\frac{1}{a+c-b}\right)+2\left(\frac{1}{b+c-a}+\frac{1}{a+b-c}\right)\)
\(+3\left(\frac{1}{a+c-b}+\frac{1}{a+b-c}\right)\)
\(\ge\frac{2}{c}+\frac{4}{b}+\frac{6}{c}\) (Do a,b,c là 3 cạnh của tam giác nên:\(\hept{\begin{cases}a+b-c>0\\a+c-b>0\\c+b-a>0\end{cases}}\)
\(=\frac{6}{a}+2a\ge4\sqrt{3}\left(cosi\right)\left(a>0\right)\)
Dấu = xảy ra khi:
\(a=b=c=\sqrt{3}\)