cho ba số x, y, z nguyên dương thỏa mãn :xy + yz + xz = 5
cmr 3x^2 + 3y^2 + z^2 ≥ 10
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NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 12 2021
\(x+\sqrt{3x+yz}=x+\sqrt{x\left(x+y+z\right)+yz}=x+\sqrt{\left(x+y\right)\left(z+x\right)}\ge x+\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}\)
\(=x+\sqrt{xz}+\sqrt{xy}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+zx}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế với vế ta có đpcm
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 4 2022
\(\sqrt{3x+yz}=\sqrt{x\left(x+y+z\right)+yz}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
\(\Rightarrow\dfrac{x}{x+\sqrt{3x+yz}}\le\dfrac{x}{x+\sqrt{xy}+\sqrt{xz}}=\dfrac{\sqrt{x}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Tương tự:
\(\dfrac{y}{y+\sqrt{3y+xz}}\le\dfrac{\sqrt{y}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\) ; \(\dfrac{z}{z+\sqrt{3z+xy}}\le\dfrac{\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\)
Cộng vế:
\(VT\le\dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
21 tháng 9 2018
\(3=x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\)
\(\Rightarrow xyz\le1\)
\(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\le\frac{x^2+1+1}{3}+\frac{y^2+1+1}{3}+\frac{z^2+1+1}{3}=3\)
Ta co:
\(A=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x\sqrt[3]{x}}{\sqrt[3]{xyz}}+\frac{y\sqrt[3]{y}}{\sqrt[3]{xyz}}+\frac{z\sqrt[3]{z}}{\sqrt[3]{xyz}}\)
\(\ge x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\)
\(\Rightarrow3A\ge3\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\ge\left(x\sqrt[3]{x}+y\sqrt[3]{y}+z\sqrt[3]{z}\right)\left(\sqrt[3]{x^2}+\sqrt[3]{y^2}+\sqrt[3]{z^2}\right)\)
\(\ge\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow A\ge xy+yz+zx\)
KN
25 tháng 5 2020
Áp dụng BĐT Cauchy - Schwarz, ta có: \(3\left(x^2+y^2+z^2\right)=\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3=x^2+y^2+z^2\)(Do \(x^2+y^2+z^2=3\))
Ta có: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{zx}}+\frac{z}{\sqrt[3]{xy}}=\frac{x}{\sqrt[3]{yz.1}}+\frac{y}{\sqrt[3]{zx.1}}+\frac{z}{\sqrt[3]{xy.1}}\)
\(\ge\frac{x}{\frac{y+z+1}{3}}+\frac{y}{\frac{z+x+1}{3}}+\frac{z}{\frac{x+y+1}{3}}\)\(=\frac{3x}{y+z+1}+\frac{3y}{z+x+1}+\frac{3z}{x+y+1}\)
\(=\frac{3x^2}{xy+zx+x}+\frac{3y^2}{yz+xy+y}+\frac{3z^2}{zx+yz+z}\)\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+\left(x+y+z\right)}\)(Theo BĐT Cauchy - Schwarz dạng Engle)
\(\ge\frac{3\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)+x^2+y^2+z^2}=\frac{3\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=3=x^2+y^2+z^2\)
\(\ge xy+yz+zx\)
Đẳng thức xảy ra khi x = y = z = 1
28 tháng 12 2019
\(\frac{x^4}{y+3z}+\frac{y+3z}{16}+\frac{1}{4}+\frac{1}{4}\ge4\sqrt[4]{\frac{x^4}{y+3z}.\frac{y+3z}{16}.\frac{1}{4}.\frac{1}{4}}=x\)
\(\Rightarrow\frac{x^4}{y+3z}\ge x-\frac{y+3z}{16}-\frac{1}{2}\)
Tương tự cho 2 BĐT còn lại :
\(\frac{y^4}{z+3x}\ge y-\frac{z+3x}{16}-\frac{1}{2};\frac{z^4}{z+3y}\ge z-\frac{x+3y}{16}-\frac{1}{2}\)
Công theo vế 3 BĐT trên ta được :
\(VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{2}\ge\frac{3}{4}.3-\frac{3}{2}=\frac{3}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!
T
29 tháng 12 2019
Cách 2:
\(VT\ge\frac{\left(x^2+y^2+z^2\right)^2}{4\left(x+y+z\right)}\ge\frac{\frac{\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2}{3}}{4\left(x+y+z\right)}\ge\frac{\left(xy+yz+zx\right)\left(x+y+z\right)}{12}\)
\(\ge\frac{\left(xy+yz+zx\right)\sqrt{3\left(xy+yz+zx\right)}}{12}\ge\frac{3}{4}\)
Đẳng thức xảy ra khi \(x=y=z=1\)
PN
26 tháng 4 2021
Áp dụng bđt phụ \(\sqrt{ \left(a+b\right)\left(c+d\right)}\ge\sqrt{ac}+\sqrt{bd}\)có
\(VT=\frac{x}{x+\sqrt{\left(x+y\right)\left(z+x\right)}}+\frac{y}{y+\sqrt{\left(y+x\right)\left(z+y\right)}}+\frac{z}{z+\sqrt{\left(z+x\right)\left(y+z\right)}}\)
\(\le\frac{x}{x+\sqrt{xz}+\sqrt{xy}}+\frac{y}{y+\sqrt{yz}+\sqrt{yx}}+\frac{z}{z+\sqrt{zx}+\sqrt{zy}}\)
\(=\frac{x}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}+\frac{y}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}+\frac{z}{\sqrt{z}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}=1\)
13 tháng 12 2015
\(VT=\frac{x^2}{x^3-xyz-2013x}+\frac{y^2}{y^3-xyz-2013y}+\frac{z^2}{z^3-xyz-2013z}\ge\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013\left(x+y+z\right)}\)
\(=\frac{\left(x+y+z\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\right]}=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\frac{1}{x+y+z}\)=VP
13 tháng 12 2015
đúng rồi ạ nhưng chỉ cần c/m đẳng thức phụ như thế này thôi ạ\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\) =>\(\frac{\left(a+b\right)2}{x+y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) hay \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\) là xong
29 tháng 4 2020
Đặt \(H=\frac{xz}{y^2+yz}+\frac{y^2}{zx+yz}+\frac{x+2z}{x+z}\)
\(=\frac{1}{\frac{y^2}{xz}+\frac{yz}{xz}}+\frac{1}{\frac{zx}{y^2}+\frac{yz}{y^2}}+\frac{x+z+z}{x+z}\)
\(=\frac{1}{\frac{y^2}{zx}+\frac{y}{x}}+\frac{1}{\frac{zx}{y^2}+\frac{z}{y}}+\frac{1}{\frac{x}{z}+1}+1\)
Đặt \(\frac{x}{y}=a;\frac{y}{z}=b\Rightarrow ab=\frac{x}{z}\ge1\)
Khi đó \(H=\frac{1}{\frac{b}{a}+\frac{1}{a}}+\frac{1}{\frac{a}{b}+\frac{1}{b}}+\frac{1}{ab+1}+1\)
\(=\frac{a}{b+1}+\frac{b}{a+b}+\frac{1}{ab+1}+1\)
Ta cần chứng minh \(U=\frac{a}{b+c}+\frac{b}{a+b}+\frac{1}{ab+1}\ge\frac{3}{2}\)
\(\Leftrightarrow\left(\frac{a}{b+1}+1\right)+\left(\frac{b}{a+1}+1\right)+\frac{1}{ab+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\frac{a+b+1}{b+1}+\frac{a+b+1}{a+1}+\frac{1}{ab+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\left(a+b+1\right)\left(\frac{1}{b+1}+\frac{1}{a+1}\right)+\frac{1}{ab+1}\ge\frac{7}{2}\)
Khi đó \(Y=\left(a+b+1\right)\left(\frac{1}{a+1}+\frac{1}{b+1}\right)+\frac{1}{ab+1}\)
\(\ge\left(a+b+1\right)\cdot\frac{4}{a+b+2}+\frac{1}{ab+1}\)
\(\ge\frac{4\left(a+b+1\right)}{a+b+2}+\frac{1}{\frac{\left(a+b\right)^2}{4}+1}\)
Đặt \(t=a+b\ge2\sqrt{ab}\ge2\)
Ta cần chứng minh \(\frac{4\left(t+1\right)}{t+2}+\frac{1}{\frac{t^2}{4}+1}\ge\frac{7}{2}\)
\(\Leftrightarrow\frac{\left(t-2\right)^3}{2\left(t+2\right)\left(t^2+4\right)}\ge0\) ( đúng )
Vậy ta có đpcm.
29 tháng 4 2020
ta có:
\(\frac{xz}{y^2+yz}+\frac{y^2}{xz+yz}+\frac{z+2z}{z+x}=\frac{\frac{xz}{yz}}{\frac{y^2}{yz}+1}+\frac{\frac{y^2}{yz}}{\frac{xz}{yz}+1}+\frac{1+\frac{2z}{x}}{1+\frac{z}{x}}\)\(=\frac{\frac{x}{y}}{\frac{y}{z}+1}+\frac{\frac{y}{z}}{\frac{x}{y}+1}+\frac{1+\frac{2z}{x}}{1+\frac{z}{x}}=\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}+\frac{1+2c^2}{1+c^2}\)
trong đó \(a^2=\frac{x}{y};b^2=\frac{y}{z};c^2=\frac{z}{x}\left(a;b;c>0\right)\)
Nhận xét rằng \(a^2\cdot b^2=\frac{x}{z}=\frac{1}{c^2}\ge1\)(do x>=z)
Xét \(\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}+\frac{c^2}{ab+1}\)\(=\frac{a^2\left(a^2+1\right)\left(ab+1\right)+b^2\left(b^2+1\right)\left(ab+1\right)-2aba^2\left(a^2+1\right)\left(b^2+1\right)}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\)
\(=\frac{ab\left(a^2-b^2\right)+\left(a-b\right)\left(a^3-b^3\right)+\left(a-b\right)^2}{\left(a^2+1\right)\left(b^2+1\right)\left(ab+1\right)}\ge0\)
Do đó: \(\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1}\ge\frac{2ab}{ab+1}=\frac{\frac{2}{c}}{\frac{1}{c}+1}=\frac{2}{1+c}\left(1\right)\)đẳng thức xảy ra <=> a=b
khi đó:
\(\frac{2}{1+c}+\frac{1+2c^2}{c^2+1}-\frac{5}{2}=\frac{2\left[2\left(1+c^2\right)+\left(1+c\right)\left(1+2c^2\right)\right]-5\left(1+c\right)\left(1+c^2\right)}{2\left(1+c\right)\left(1+c^2\right)}\)
\(=\frac{1-3c+3c^2-c^3}{2\left(1+c\right)\left(1+c^2\right)}=\frac{\left(1-c\right)^3}{2\left(1+c\right)\left(1+c^2\right)}\ge0\)(do c=<1) (2)
Từ (1) và (2) => đpcm
Đẳng thức xảy ra <=> a=b, c=1 <=> x=y=z
Lời giải:
Áp dụng BĐT AM-GM cho các số thực dương ta có:
$2x^2+\frac{z^2}{2}\geq 2xz$
$2y^2+\frac{z^2}{2}\geq 2yz$
$x^2+y^2\geq 2xy$
Cộng theo vế và thu gọn suy ra:
$3x^2+3y^2+z^2\geq 2(xy+yz+xz)=10$
(đpcm)
Dấu "=" xảy ra khi \(2x=2y=z=2\)