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Vì xyz = 1 nên ta có thể đặt \(x=\frac{a^2}{bc};y=\frac{b^2}{ac};z=\frac{c^2}{ab}\left(a,b,c>0,a^2\ne bc,b^2\ne ac,c^2\ne ab\right)\)
Khi đó bất đẳng thức tương đương với
\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge1\)
Mà ta có
\(\frac{a^4}{\left(a^2-bc\right)^2}+\frac{b^4}{\left(b^2-ac\right)^2}+\frac{c^4}{\left(c^2-ab\right)^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\)
Ta cần chứng minh
\(\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2}\ge1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\left(a^2-bc\right)^2+\left(b^2-ab\right)^2+\left(c^2-ab\right)^2\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge0\left(đúng\right)\)
Vậy ta có điều phải chứng minh
Bài 2:
Tìm GTLN: \(x^2+xy+y^2=3\Leftrightarrow xy=\left(x+y\right)^2-3\Rightarrow xy\ge-3\Rightarrow-7xy\le21\)
\(P=2\left(x^2+xy+y^2\right)-7xy\le2.3+21=27\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\xy=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{3},y=-\sqrt{3}\\x=-\sqrt{3},y=\sqrt{3}\end{cases}}\)
Tìm GTNN:
Chứng minh \(xy\le\frac{1}{2}\left(x^2+y^2\right)\Rightarrow\frac{3}{2}xy\le\frac{1}{2}\left(x^2+y^2+xy\right)\)
\(\Rightarrow\frac{3}{2}xy\le\frac{3}{2}\Rightarrow xy\le1\Rightarrow-7xy\ge-7\)
\(P=2\left(x^2+xy+y^2\right)-7xy\ge2.3-7=-1\)
Chúc bạn học tốt.
Làm bài 1 ha :)
Áp dụng BĐT Cô si ta có:
\(\left(1-x^3\right)+\left(1-y^3\right)+\left(1-z^3\right)\ge3\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
\(\Leftrightarrow\frac{3-\left(x^3+y^3+z^3\right)}{3}\ge\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)
Mặt khác:\(\frac{3-\left(x^3+y^3+z^3\right)}{3}\le\frac{3-3xyz}{3}=1-xyz\)
Khi đó:
\(\left(1-xyz\right)^3\ge\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)\)
Giống Holder ghê vậy ta :D
Ta có: \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\left(1+\frac{x}{y}+\frac{y}{z}+\frac{x}{z}\right)\left(1+\frac{z}{x}\right)=2+\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{x}+\frac{x}{z}\)
\(=2+\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\right)\)
Ta chứng minh bất đẳng thức :
\(\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\right)\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Vì x, y, z đóng vai trò như nhau nên ta chứng minh bất đẳng thức phụ:
\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}\)
Xét:
\(3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)=\left(\frac{2x}{y}+\frac{y}{z}\right)+\left(\frac{2y}{z}+\frac{z}{x}\right)+\left(\frac{2z}{x}+\frac{x}{y}\right)\)
Áp dụng BĐT AM-GM ta có:
\(\frac{2x}{y}+\frac{y}{z}=\frac{x}{y}+\frac{x}{y}+\frac{y}{z}\ge3\sqrt[3]{\frac{x.x.y}{y.y.z}}=3\sqrt[3]{\frac{x.x.x}{xyz}}=3\frac{x}{\sqrt[3]{xyz}}\)
Tương tự như thế ta có:
\(3\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)\ge3.\frac{x}{\sqrt[3]{xyz}}+3\frac{y}{\sqrt[3]{xyz}}+3\frac{z}{\sqrt[3]{xyz}}\)
\(\Rightarrow\)\(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}\)
Như vậy:
\(\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}\right)\ge\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
=> \(\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\ge2+\frac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Dấu "=" khi x=y=z
Câu hỏi của Incursion_03 - Toán lớp 9 - Học toán với OnlineMath
Ta có:
\(\frac{x}{1+x^2}+\frac{18y}{1+y^2}+\frac{4z}{1+z^2}=xyz\left(\frac{1}{yz\left(1+x^2\right)}+\frac{18}{xz\left(1+y^2\right)}+\frac{4}{xy\left(1+z^2\right)}\right)\)
\(=xyz\left(\frac{1}{yz+x\left(x+y+z\right)}+\frac{18}{xz+y\left(x+y+z\right)}+\frac{4}{xy+z\left(x+y+z\right)}\right)\)
\(=xyz\left(\frac{1}{\left(x+y\right).\left(x+z\right)}+\frac{18}{\left(y+x\right).\left(y+z\right)}+\frac{4}{\left(z+x\right).\left(z+y\right)}\right)\)
\(=xyz.\frac{\left(z+y\right)+18.\left(x+z\right)+4\left(x+y\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)
\(=\frac{xyz\left(22x+5y+19z\right)}{\left(x+y\right).\left(y+z\right).\left(z+x\right)}\)(đpcm)
\(P=\frac{1}{x^2+y^2+z^2}+\frac{2009}{xy+yz+zx}=\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}+\frac{2007}{xy+yz+zx}\)
\(P\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{2007}{\frac{1}{3}\left(x+y+z\right)^2}\)
\(P\ge\frac{9}{\left(x+y+z\right)^2}+\frac{6021}{\left(x+y+z\right)^2}=\frac{6030}{\left(x+y+z\right)^2}\ge\frac{6030}{3^2}=670\)
Dấu "=" xảy ra khi \(x=y=z=1\)
Áp dụng BĐT Côsi dưới dạng engel, ta có:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\)
⇒\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right).\frac{9}{x+y+z}\) = 9
Dấu "=" xảy ra ⇔ x = y = z
Theo bài ra ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\Rightarrow x+y+z=xyz\)
Do:\(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)
Tương tự: \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\);
\(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(z+y\right)\left(x+y\right)}\)
\(A=\sqrt{\frac{x^2}{yz\left(1+x^2\right)}}+\sqrt{\frac{y^2}{zx\left(1+y^2\right)}}+\sqrt{\frac{z^2}{xy\left(1+z^2\right)}}\)
\(A=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)
Áp dụng bất đẳng thức Cô si \(\frac{a+b}{2}\ge\sqrt{ab}\), dấu "=" xảy ra khi \(a=b\)
Ta có \(\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\);
\(\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\);
\(\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\le\frac{1}{2}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\)
\(A\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+z}+\frac{y}{y+x}+\frac{z}{y+z}+\frac{z}{x+z}\right)=\frac{3}{2}\)
Vậy \(A\le\frac{3}{2}\). Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)
M giải thích cho t chỗ sao mà \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\) đc vậy?
Với cả từ dòng này xuống dòng này nữa.
Sao mà tin đc dấu " = " xảy ra khi nào vậy?
Ta có : \(27xyz\le\left(x+y+z\right)^3\)
<=> \(\left(x+y+z\right)^3-27xyz\ge0\)
<=> (x + y)3 + 3(x + y)z(x + y + z) + z3 - 27xyz \(\ge0\)
=> x3 + y3 + 3xy(x + y) + 3(x + y)z(x + y + z) + z3 - 27xyz \(\ge\)0
<=> (x3 + y3 + z3) + 3(x + y)[xy + z(x + y + z)] - 27xyz \(\ge0\)
<=> (x3 + y3 + z3) + 3(x + y)(y + z)(z + x) - 27xyz \(\ge0\)
mà x + y \(\ge2\sqrt{xy}\)
Thật vậy x + y \(\ge2\sqrt{xy}\)
=> (x + y)2 \(\ge\)4xy
<=> x2 - 2xy + y2 \(\ge\) 0
<=> (x - y)2 \(\ge\)0 (đúng \(\forall x;y>0\))
Tương tự ta được y + z \(\ge2\sqrt{yz}\)
z + x \(\ge2\sqrt{xz}\)
Khi đó 3(x + y)(y + z)(z + x) \(\ge3.2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}=24xyz\)(dấu "=" xảy ra khi x = y = z)
=> (x3 + y3 + z3) + 3(x + y)(y + z)(z + x) - 27xyz \(\ge0\)
<=> (x3 + y3 + z3) + 24xyz - 27xyz \(\ge0\)
<=> x3 + y3 + z3 - 3xyz \(\ge0\)
<=> (x + y + z)[(x - y)2 + (y - z)2 + (z - x)2] \(\ge\)0 (đúng)
=> ĐPCM