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\(\frac{3}{x\sqrt{x}}=3\sqrt[3]{y^2z^2t^2}\le yz+zt+ty\)
\(\Sigma\frac{1}{x^3\left(yz+zt+ty\right)}\ge\Sigma\frac{1}{\frac{3x^3}{x\sqrt{x}}}=\Sigma\frac{\sqrt{x}}{3x^2}\ge\frac{4}{3}\sqrt[4]{\frac{\sqrt{xyzt}}{\left(xyzt\right)^2}}=\frac{4}{3}\)
Câu hỏi của Ryan Park - Toán lớp 9 - Học toán với OnlineMath
Chứng minh đc:
\(\frac{1}{x^3\left(yz+zt+ty\right)}+\frac{1}{y^3\left(xz+zt+tx\right)}+\frac{1}{z^3\left(xy+yt+tx\right)}+\frac{1}{t^3\left(xy+yz+zx\right)}\)
\(\ge\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\)
\(\ge\frac{4}{3}.\sqrt[4]{\frac{1}{xyzt}}=\frac{4}{3}\)
Áp dụng Bất đẳng thức AM-GM cho 4 số dương :
\(\Rightarrow2x+xy+z+yzt\ge4\sqrt[4]{2x^2y^2z^2t}\)
\(\Rightarrow1\ge4\sqrt[4]{2x^2y^2z^2t}\Rightarrow1\ge512.x^2y^2z^2t\Rightarrow x^2y^2z^2t\le\dfrac{1}{512}\)
=> MaxI=\(\dfrac{1}{152}\) khi \(\left\{{}\begin{matrix}x=\dfrac{1}{8}\\y=2\\z=\dfrac{1}{4}\\t=\dfrac{1}{2}\end{matrix}\right.\)
Hà Nam Phan Đình cho tớ hỏi BĐT AM-GM là BĐT gì vậy? và lớp mấy được hok vậy ạ?
Ta đặt: \(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c;\frac{1}{t}=d\) ( a, b, c, d >0 )
Khi đó ta cần chứng minh:
\(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)
\(VT=\frac{a^3}{\frac{b+c+d}{bcd}}+\frac{b^3}{\frac{a+c+d}{acd}}+\frac{c^3}{\frac{a+b+d}{abd}}+\frac{d^3}{\frac{a+b+c}{abc}}\)
\(=\frac{a^3}{\frac{a\left(b+c+d\right)}{abcd}}+\frac{b^3}{\frac{b\left(a+c+d\right)}{abcd}}+\frac{c^3}{\frac{c\left(a+b+d\right)}{abcd}}+\frac{d^3}{\frac{d\left(a+b+c\right)}{abcd}}\)
\(=\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{3\left(a+b+c+d\right)}=\frac{a+b+c+d}{3}=VP\)
Vậy ta đã chứng minh được
\(\frac{a^3}{\frac{1}{bc}+\frac{1}{cd}+\frac{1}{db}}+\frac{b^3}{\frac{1}{ac}+\frac{1}{cd}+\frac{1}{da}}+\frac{c^3}{\frac{1}{ab}+\frac{1}{bd}+\frac{1}{da}}+\frac{d^3}{\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}}\ge\frac{1}{3}\left(a+b+c+d\right)\)
Dấu "=" xảy ra <=> a = b = c = d
Vậy :
\(\frac{1}{x^3\left(yz+zt+ty\right)}+\frac{1}{y^3\left(xz+zt+tx\right)}+\frac{1}{z^3\left(xy+yt+tx\right)}+\frac{1}{t^3\left(xy+yz+zx\right)}\ge\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{t}\right)\)
Dấu "=" xảy ra <=> x = y = z = t = 1
Ta có:
\(4A=\frac{\left(x+y+z+t\right)^2\left(x+y+z\right)\left(x+y\right)}{xyzt}\)
\(\ge\frac{4\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{xyzt}\)
\(=\frac{4\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{16\left(x+y\right)z\left(x+y\right)}{xyz}\)
\(=\frac{16\left(x+y\right)^2}{xy}\ge\frac{64xy}{xy}=64\)
\(\Rightarrow A\ge16\)
Đấu = xảy ra khi \(t=2z=4x=4y=1\)
x;y;z;t >0 áp dụng bất đẳng thức Cô-si cho 2 số dương ta có :
=\(x+y\ge2\sqrt{xy}\)
=\(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)
=\(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\)
nhân các vế tương ứng ta có:
\(\left(x+y\right)\left(x+y+z\right)\left(x+y+z+t\right)\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)
mà x+y+z+t=2
\(\left(x+y\right)\left(x+y+z\right)2\ge8\sqrt{xyzt\left(x+y\right)\left(x+y+z\right)}\)
=\(\sqrt{\left(x+y\right)\left(x+y+z\right)}\ge4\sqrt{xyzt}\)
=\(\left(x+y\right)\left(x+y+z\right)\ge16xyzt\)
\(\Rightarrow B=\frac{\left(x+y\right)\left(x+y+z\right)}{xyzt}\ge\frac{16xyzt}{xyzt}=16\)
vậy minB=16 khi\(\hept{\begin{cases}x=y\\x+y=z\\x+y+z=t\end{cases}};x+y+z+t=2\Rightarrow x=y=0.25;z=0.5;t=1\)
\(A=\frac{2^2\left(x+y+z\right)\left(x+y\right)}{4xyzt}=\frac{\left(x+y+z+t\right)^2\left(x+y+z\right)\left(x+y\right)}{4xyzt}\)
\(A\ge\frac{4\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{4xyzt}=\frac{\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{4\left(x+y\right)^2z\left(x+y\right)}{xyz}\)
\(A\ge\frac{4\left(x+y\right)^2}{xy}\ge\frac{16xy}{xy}=16\)
\(A_{min}=16\) khi \(\left\{{}\begin{matrix}x+y+z+t=2\\x+y+z=t\\x+y=z\\x=y\end{matrix}\right.\) \(\Rightarrow\left(x;y;z;t\right)=...\)
Ta có:
\(x+y+z+t=2\)
\(\Rightarrow\left[\left(x+y+z\right)+t\right]^2=4\)
Vì \(x,y,z,t>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(\left(x+y+z\right)+t\ge2\sqrt{\left(x+y+z\right)t}\)
\(\Leftrightarrow\left[\left(x+y+z\right)+t\right]^2\ge4\left(x+y+z\right)t\)
\(\Leftrightarrow4\ge4\left(x+y+z\right)t\)(vì \(\left[\left(x+y+z\right)+t\right]^2=4\))
\(\Leftrightarrow\left(x+y+z\right)t\le1\left(1\right)\)
Ta có:
\(P=\frac{\left(x+y+z\right)\left(x+y\right)}{xyzt}=\frac{1.\left(x+y+z\right)\left(x+y\right)}{xyzt}\)
\(\Leftrightarrow P\ge\frac{\left(x+y+z\right)t\left(x+y+z\right)\left(x+y\right)}{xyzt}\)(vì (1))
\(\Leftrightarrow P\ge\frac{\left(x+y+z\right)^2\left(x+y\right)}{xyz}\left(2\right)\)
Đặt \(\frac{\left(x+y+z\right)^2\left(x+y\right)}{xyz}=A\)thì \(P\ge A\)
Vì \(x,y,z>0\)nên áp dụng bất đẳng thúc Cô-si cho 2 số dương, ta được:
\(\left(x+y\right)+z\ge2\sqrt{\left(x+y\right)z}\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge4\left(x+y\right)z\)
Do đó:
\(A=\frac{\left(x+y+z\right)^2\left(x+y\right)}{xyz}\ge\frac{4\left(x+y\right)z\left(x+y\right)}{xyz}\)
\(\Leftrightarrow A\ge\frac{4\left(x+y\right)^2}{xy}\left(3\right)\)
Từ (2) và (3), ta được:
\(P\ge\frac{4\left(x+y\right)^2}{xy}\left(4\right)\)
Vì \(x,y>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:
\(x+y\ge2\sqrt{xy}\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow4\left(x+y\right)^2\ge16xy\)
\(\Leftrightarrow\frac{4\left(x+y\right)^2}{xy}\ge\frac{16xy}{xy}=16\left(5\right)\)
Từ (4) và (5), ta được:
\(P\ge16\)
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}x=y>0\\x+y=z>0\\x+y+z=t>0\end{cases}}\)
Mà \(x+y+z+t=2\)nên:
\(\Leftrightarrow\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\\t=1\end{cases}}\)
Vậy \(minP=16\Leftrightarrow x=y=\frac{1}{4};z=\frac{1}{2};t=1\)