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Đặt \(B=xy=2013-A\) thế vô cái cần tìm thì được
\(5x^2+\frac{y^2}{4}+\frac{1}{4x^2}=\frac{5}{2}\)
\(\Leftrightarrow x^2y^2+20x^4-10x^2+1=0\)
\(\Leftrightarrow20x^4-10x^2+1+B^2=0\)
\(\Leftrightarrow B^2=\frac{1}{4}-\left(\sqrt{20}x^2-\frac{\sqrt{5}}{2}\right)^2\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{2}\le B\le\frac{1}{2}\)
\(\Leftrightarrow-\frac{1}{2}\le2013-A\le\frac{1}{2}\)
\(\Leftrightarrow2012,3\le A\le2013,5\)
\(4=a^2+\frac{1}{a^2}+a^2+\frac{b^2}{4}\ge2\sqrt{\frac{a^2}{a^2}}+2\sqrt{\frac{a^2b^2}{4}}\)
\(\Rightarrow4\ge2+ab\Rightarrow ab\le2\)
\(\Rightarrow P_{max}=2\) khi \(\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}a=-1\\b=-2\end{matrix}\right.\)
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 \(a=b=\frac{1}{2}\)
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)
\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có :
\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)
\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)
\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)
\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)
Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)
PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))
nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm
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: \(m=n=p=1\Leftrightarrow a=b=c=1\)
Ta có: \(2a^2+\frac{b^2}{4}+\frac{1}{a^2}=4\Rightarrow8a^4+a^2b^2+4=16a^2\Rightarrow a^2b^2=-8a^4+16a^2-4=-8\left(a^4-2a^2+1\right)+4=-8\left(a^2-1\right)^2+4\le4\)\(\Rightarrow\left|ab\right|\le2\Rightarrow-2\le ab\le2\)
Vậy MaxS = 2023 khi ab = 2 và a2 = 1 do đó \(\left(a,b\right)\in\left\{\left(-1;-2\right);\left(1;2\right)\right\}\)
MinS = 2019 khi ab = -2 và a2 = 1 do đó \(\left(a,b\right)\in\left\{\left(-1;2\right);\left(1;-2\right)\right\}\)