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\(\frac{x}{x^2-yz+2013}+\frac{y}{y^2-zx+2013}+\frac{z}{z^2-xy+2013}\)
\(=\frac{1}{\frac{x^2-yz+2013}{x}}+\frac{1}{\frac{y^2-zx+2013}{y}}+\frac{1}{\frac{z^2-xy+2013}{z}}\)
\(=\frac{1}{x+3y+3z+\frac{2yz}{x}}+\frac{1}{y+3z+3x+\frac{2xz}{y}}+\frac{1}{z+3x+3y+\frac{2xy}{z}}\)
\(\ge\frac{9}{7\left(x+y+z\right)+2xyz\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\ge\frac{9}{7\left(x+y+z\right)+2xyz\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}=\)
\(=\frac{9}{7\left(x+y+z\right)+2xyz.\frac{1}{xyz}.\left(x+y+z\right)}=\frac{9}{9\left(x+y+z\right)}=\frac{1}{x+y+z}\)
Ta có đpcm
bó tay rùi bạn !!!! ~_~
65756578687696453724756545345363637635754754695622534434
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{xy}{\sqrt{z+xy}}=\frac{xy}{\sqrt{z\left(x+y+z\right)+xy}}=\frac{xy}{\sqrt{xz+yz+z^2+xy}}\)
\(=\frac{xy}{\sqrt{\left(x+z\right)\left(y+z\right)}}\le\frac{1}{2}\left(\frac{xy}{x+z}+\frac{xy}{y+z}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\frac{yz}{\sqrt{x+yz}}\le\frac{1}{2}\left(\frac{yz}{x+y}+\frac{yz}{x+z}\right);\frac{xz}{\sqrt{y+xz}}\le\frac{1}{2}\left(\frac{xz}{y+z}+\frac{xz}{x+y}\right)\)
Cộng theo vế các BĐT trên ta có:
\(P\le\frac{1}{2}\left(\frac{xy+yz}{x+z}+\frac{yz+xz}{x+y}+\frac{xy+xz}{y+z}\right)\)
\(=\frac{1}{2}\left(\frac{y\left(x+z\right)}{x+z}+\frac{z\left(x+y\right)}{x+y}+\frac{x\left(y+z\right)}{y+z}\right)\)
\(=\frac{1}{2}\left(x+y+z\right)=\frac{1}{2}\left(x+y+z=1\right)\)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)
\(P^2=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.\left(\frac{xy.yz}{zx}+\frac{yz.zx}{xy}+\frac{zx.xy}{zy}\right)\)
\(=\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}+2.2016\)
Áp dụng BĐT Cauchy:\(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}\ge2\sqrt{\frac{x^2y^2}{z^2}.\frac{y^2z^2}{x^2}}=2y^2\)
\(\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2\sqrt{\frac{y^2z^2}{x^2}.\frac{z^2x^2}{y^2}}=2z^2\)
\(\frac{z^2x^2}{y^2}+\frac{x^2y^2}{z^2}\ge2\sqrt{\frac{x^2z^2}{y^2}.\frac{x^2y^2}{z^2}}=2x^2\)
Cộng theo vế ta được:\(2\left(\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\right)\ge2x^2+2y^2+2z^2=2.2016\)
\(\Rightarrow\frac{x^2y^2}{z^2}+\frac{y^2z^2}{x^2}+\frac{z^2x^2}{y^2}\ge2016\)
\(\Rightarrow P^2\ge2016+2016.2=6048\Rightarrow P\ge\sqrt{6048}=12\sqrt{42}\)
Nên GTNN của P là \(12\sqrt{42}\) đạt được khi \(x=y=z=\sqrt{\frac{2016}{3}}=4\sqrt{42}\)
a) ĐKXĐ: \(x;y>0\)
Ta có:\(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)
\(\Rightarrow\frac{4y}{4xy}+\frac{4x}{4xy}=\frac{xy}{4xy}\)
\(\Rightarrow4x+4y-xy=0\)
\(\Rightarrow x\left(4-y\right)=-4y\)
\(\Rightarrow x=\frac{-4y}{4-y}=\frac{-4\left(y-4\right)-16}{-\left(y-4\right)}\)
\(\Rightarrow x=4-\frac{16}{4-y}\)
Để x nguyên dương =>\(\hept{\begin{cases}\frac{16}{4-y}< 0\\\left(4-y\right)\inƯ\left(16\right)\end{cases}}\)
\(\Rightarrow4-y\in\left\{\pm1;\pm2;\pm4;\pm8;\pm16\right\}\)
Tìm nốt y và thay vào tìm ra x
a/ \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\)
Không mất tính tổng quát giả sử: \(x\ge y\)
\(\frac{1}{4}=\frac{1}{x}+\frac{1}{y}\le\frac{2}{y}\)
\(\Leftrightarrow0< y\le8\)
\(\Rightarrow y=\left\{1;2;3;4;5;6;7;8\right\}\)làm nốt
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
\(=\frac{1}{\sqrt{\frac{1}{2}\left(x-y\right)^2+\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y-z\right)^2+\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z-x\right)^2+\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{1}{\sqrt{\frac{1}{2}\left(x^2+y^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(y^2+z^2\right)}}+\frac{1}{\sqrt{\frac{1}{2}\left(z^2+x^2\right)}}\)
\(\le\frac{2}{x+y}+\frac{2}{y+z}+\frac{2}{z+x}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Dat \(\left(a,b,c\right)=\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\left(a,b,c>0,abc=1\right)\)
Ta co \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Rightarrow\frac{3}{ab+bc+ca}\ge\frac{9}{\left(a+b+c\right)^2}\left(1\right)\)
BDT phu \(1+\frac{3}{ab+bc+ca}\ge\frac{6}{a+b+c}\left(2\right)\)
Do (1) nen (2) tuong duong voi
\(1+\frac{9}{\left(a+b+c\right)^2}\ge\frac{6}{a+b+c}\Leftrightarrow\left(1-\frac{3}{a+b+c}\right)^2\ge0\left(dung\right)\)
Suy ra (2) duoc chung minh
Do \(abc=1\Rightarrow\hept{\begin{cases}ab=\frac{1}{xy}=\frac{xyz}{xy}=z\\bc=x\\ca=y\end{cases}}\)
nen (2) tuong duong \(1+\frac{3}{x+y+z}\ge\frac{6}{xy+yz+zx}\)
=> \(\frac{1}{x+y+z}\ge\frac{1}{3}\left(\frac{6}{x+y+z}-1\right)=\frac{2}{x+y+z}-\frac{1}{3}\)
Suy ra \(P\ge\frac{2}{x+y+z}-\frac{1}{3}-\frac{2}{x+y+z}=-\frac{1}{3}\)
Dau = xay ra khi x=y=z=1
Ta có: \(xy+yz+zx=xyz\Leftrightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\)ta có: \(a,b,c>0;a+b+c=1\)do đó 0<a,b,c<1
\(P=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+6\left(ab+bc+ca\right)\)
\(=\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}+2\left(a+b+c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\left(\frac{b^2}{a}-2b+a\right)+\left(\frac{c^2}{b}-2c+b\right)+\left(\frac{a^2}{c}-2a+c\right)-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(a-b\right)^2}{a}+\frac{\left(b-c\right)^2}{b}+\frac{\left(c-a\right)^2}{c}-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+3\)
\(=\frac{\left(1-a\right)\left(a-b\right)^2}{a}+\frac{\left(1-b\right)\left(b-c\right)^2}{b}+\frac{\left(1-c\right)\left(c-a\right)^2}{c}+3\ge3\)
Vậy GTNN của P=3