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a/ \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=\left(xy-\frac{1}{xy}\right)^2+4\ge4\)
Suy ra Min M = 4 . Dấu "=" xảy ra khi x=y=1/2
b/ Đề đúng phải là \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{3}{2}\)
Ta có \(6=\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\ge\frac{9}{2\left(x+y+z\right)}\Rightarrow x+y+z\ge\frac{3}{4}\)
Lại có \(\frac{1}{3x+3y+2z}+\frac{1}{3x+2y+3z}+\frac{1}{2x+3y+3z}\ge\frac{9}{8\left(x+y+z\right)}\ge\frac{9}{8.\frac{3}{4}}=\frac{3}{2}\)
\(BDT\Leftrightarrow\text{∑}\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)\ge\frac{21}{2}\)
Mà \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\). Vậy ta cần chứng minh
\(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{z^2}{x^2}+\frac{x^2}{z^2}\ge\frac{17}{2}\)
\(\Leftrightarrow\frac{y^2}{z^2}+\frac{x^2}{z^2}\ge\frac{1}{2}\left(\frac{x}{z}+\frac{y}{z}\right)^2\)
\(\Leftrightarrow\frac{z^2}{y^2}+\frac{z^2}{x^2}\ge\frac{1}{2}\left(\frac{4z}{x+y}\right)^2\)
Đặt \(a=\frac{z}{x+y}\ge1\), ta chứng minh \(\frac{1}{2a^2}+8a^2\ge\frac{17}{2}\)
Dễ thấy BĐT này đúng. Vậy ta có đpcm
1) BĐT chứng minh ⇔∑(x2y2+y2x2)≥212⇔∑(x2y2+y2x2)≥212
Ta có x2y2+y2x2≥2x2y2+y2x2≥2
Ta sẽ đi chứng minh y2z2+z2y2+z2x2+x2z2≥172y2z2+z2y2+z2x2+x2z2≥172
Ta có y2z2+x2z2≥12(xz+yz)2y2z2+x2z2≥12(xz+yz)2
z2y2+z2x2≥12(4zx+y)2z2y2+z2x2≥12(4zx+y)2
Đặt a=zx+y≥1a=zx+y≥1
Ta sẽ chứng minh 12a2+8a2≥17212a2+8a2≥172
Dễ thấy bđt này đúng suy ra đpcm
\(BDT\Leftrightarrow\text{∑}\left(\frac{x^2}{y^2}+\frac{y^2}{x^2}\right)\ge\frac{21}{2}\)
Mà \(\frac{x^2}{y^2}+\frac{y^2}{x^2}\ge2\)(dùng AM-GM giải quyết chỗ này)
Vậy ta cần chứng minh \(\frac{y^2}{z^2}+\frac{z^2}{y^2}+\frac{z^2}{x^2}+\frac{x^2}{z^2}\ge\frac{17}{2}\)
\(\Leftrightarrow\frac{y^2}{z^2}+\frac{x^2}{z^2}\ge\frac{1}{2}\left(\frac{x}{z}+\frac{y}{z}\right)^2\)
\(\Leftrightarrow\frac{z^2}{y^2}+\frac{z^2}{x^2}\ge\frac{1}{2}\left(\frac{4z}{x+y}\right)^2\)
Đặt \(a=\frac{z}{x+y}\ge1\),ta chứng minh \(\frac{1}{2a^2}+8a^2\ge\frac{17}{2}\)
Dễ thấy BĐT này đúng.Vậy ta có đpcm
Câu 1:
\(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)=x^2y^2+\frac{1}{x^2y^2}+2=x^2y^2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}+2\)
\(\ge\frac{1}{8}+2+\frac{255}{256x^2y^2}\)
Ta lại có: \(1=x+y\ge2\sqrt{xy}\Leftrightarrow1\ge16x^2y^2\)
\(\Rightarrow M\ge\frac{17}{8}+\frac{255}{16}=\frac{289}{16}\)
Dấu = xảy ra khi x=y=1/2
Áp dụng BDT Cauchy-Schwarz: \(\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\right)\ge\frac{1}{3x+3y+2z}\)
CMTT rồi cộng vế với vế ta có.\(VT\le\frac{1}{16}\cdot4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)
Dấu = xảy ra khi x=y=z=1
1)
+) Ta có
\(\left(a-b\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2-2ab\ge0\)
\(\Leftrightarrow a^2+b^2\ge2ab\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge a^2+b^2+2ab\)
\(\Leftrightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Leftrightarrow a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\) ( đpcm )
+ ) Theo phần trên
\(a^2+b^2\ge2ab\)
\(\Leftrightarrow a^2+b^2+2ab\ge4ab\)
\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)
\(\Leftrightarrow ab\le\frac{1}{4}\left(a+b\right)^2\) ( đpcm )
2,
Ta có: \(5\left(x^2+y^2+z^2\right)-9x\left(y+z\right)-18yz=0\Leftrightarrow5x^2-9x\left(y+z\right)+5\left(y+z\right)^2=28yz\le7\left(y+z\right)^2\)\(\Leftrightarrow5x^2-9x\left(y+z\right)-2\left(y+z\right)^2\le0\Leftrightarrow5\left(\frac{x}{y+z}\right)^2-9.\frac{x}{y+z}-2\le0\)\(\Leftrightarrow\left(5.\frac{x}{y+z}+1\right)\left(\frac{x}{y+z}-2\right)\le0\Leftrightarrow\frac{x}{y+z}\le2\)(Do \(5.\frac{x}{y+z}+1>0\forall x,y,z>0\))
\(\Rightarrow E=\frac{2x-y-z}{y+z}=2.\frac{x}{y+z}-1\le2.2-1=3\)
Đẳng thức xảy ra khi \(y=z=\frac{x}{4}\)
Ta có \(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\frac{\left(yz+1\right)^2}{z^2}}{\frac{zx+1}{x}}+\frac{\frac{\left(zx+1\right)^2}{x^2}}{\frac{xy+1}{y}}+\frac{\frac{\left(xy+1\right)^2}{y^2}}{\frac{yz+1}{z}}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT \(\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\frac{a_3^2}{b_3}\ge\frac{\left(a_1+a_2+a_3\right)^2}{b_1+b_2+b_3}\)
Dấu "=" xảy ra khi \(\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{c_3}\)
\(P=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}\)
\(P\ge a+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng BĐT: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
=> \(P\ge x+y+z+\frac{9}{x+y+z}=\left[x+y+z+\frac{9}{4\left(x+y+z\right)}\right]+\frac{27}{4\left(x+y+z\right)}\)
Ta có: \(x+y+z+\frac{9}{4\left(x+y+z\right)}\ge2\sqrt{\frac{9}{4}}=3;\frac{27}{4\left(x+y+z\right)}=\frac{27}{4\cdot\frac{3}{2}}=\frac{9}{2}\)
=> \(P\ge3+\frac{9}{2}=\frac{15}{2}\).
Dấu "=" xảy ra <=> x=y=z=\(\frac{1}{2}\)
Vậy MinP=\(\frac{15}{2}\)đạt được khi x=y=z=\(\frac{1}{2}\)
Ta có:
\(P=\frac{x\left(yz+1\right)^2}{z^2\left(zx+1\right)}+\frac{y\left(zx+1\right)^2}{x^2\left(xy+1\right)}+\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}\)
\(=\frac{\left(\frac{yz+1}{z}\right)^2}{\left(\frac{zx+1}{x}\right)}+\frac{\left(\frac{zx+1}{x}\right)^2}{\left(\frac{xy+1}{y}\right)}+\frac{\left(\frac{xy+1}{y}\right)^2}{\left(\frac{yz+1}{z}\right)}\)
\(=\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)
Áp dụng BĐT Bunhiacopxki dạng phân thức, ta có:
\(\frac{\left(y+\frac{1}{z}\right)^2}{z+\frac{1}{x}}+\frac{\left(z+\frac{1}{x}\right)^2}{x+\frac{1}{y}}+\frac{\left(x+\frac{1}{y}\right)^2}{y+\frac{1}{z}}\)\(\ge\frac{\left(x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}{x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\ge\left(x+y+z\right)+\frac{9}{x+y+z}=\left(x+y+z\right)+\frac{9}{4\left(x+y+z\right)}\)
\(+\frac{27}{4\left(x+y+z\right)}\ge2\sqrt{\left(x+y+z\right).\frac{9}{4\left(x+y+z\right)}}+\frac{27}{4.\frac{3}{2}}=\frac{15}{2}\)(Áp dụng BĐT Cô - si cho 2 số không âm)
Đẳng thức xảy ra khi \(x=y=z=\frac{1}{2}\)
\(\left[2\left(x-1\right)-3\left(y+2\right)+4\left(z-3\right)\right]^2\le\left(2^2+3^2+4^2\right)\left[\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2\right]\)
\(\Rightarrow\left(2x-3y+4z-20\right)^2\le29\)
\(\Rightarrow\left|2x-3y+4z-20\right|\le\sqrt{29}\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-1\right)^2+\left(y+2\right)^2+\left(z-3\right)^2=1\\\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}\end{matrix}\right.\)