Cho x, y, z là các số thực lớn hơn -1 . Chứng minh: \(\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\ge2\)
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2 tháng 4 2020
Ta có :
\(A=\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi \(x=y=z=1\)
2 tháng 4 2020
Ta có :
\(\frac{1+x^2}{1+y+z^2}+\frac{1+y^2}{1+z+x^2}+\frac{1+z^2}{1+x+y^2}\)
\(\Rightarrow A=\frac{1+z+x^2}{1+y+z^2}+\frac{1+x+y^2}{1+z+x^2}+\frac{1+y+z^2}{1+x+y^2}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3\sqrt[3]{\frac{1+z+x^2}{1+y+z^2}.\frac{1+x+y^2}{1+z+x^2}.\frac{1+y+z^2}{1+x+y^2}}\)
\(-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{1+y+z^2}+\frac{x}{1+z+x^2}+\frac{y}{1+x+y^2}\right)\)
\(\Rightarrow A\ge3-\left(\frac{z}{y+2z}+\frac{x}{z+2x}+\frac{y}{x+2y}\right)\)
\(\Rightarrow A\ge3-\left(\frac{1}{2}-\frac{y}{2\left(y+2z\right)}+\frac{1}{2}-\frac{z}{2\left(z+2x\right)}+\frac{1}{2}-\frac{x}{2\left(x+2y\right)}\right)\)
\(\Rightarrow A\ge3-\frac{3}{2}+\frac{1}{2}\left(\frac{y}{y+2z}+\frac{z}{z+2x}+\frac{x}{x+2y}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{y^2}{y^2+2yz}+\frac{z^2}{z^2+2xz}+\frac{x^2}{x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{y^2+2yz+z^2+2xz+x^2+2xy}\right)\)
\(\Rightarrow A\ge\frac{3}{2}+\frac{1}{2}\left(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}\right)\)
\(\Rightarrow A\ge2\)
Dấu " = " xảy ra khi x=y=z=1
TN
18 tháng 9 2016
Bài 1: \(T=\sqrt{\frac{x^3}{x^3+8y^3}}+\sqrt{\frac{4y^3}{y^3+\left(x+y\right)^3}}\)
\(=\frac{x^2}{\sqrt{x\left(x^3+8y^3\right)}}+\frac{2y^2}{\sqrt{y\left[y^3+\left(x+y\right)^3\right]}}\)
\(=\frac{x^2}{\sqrt{\left(x^2+2xy\right)\left(x^2-2xy+4y^2\right)}}+\frac{2y^2}{\sqrt{\left(xy+2y^2\right)\left(x^2+xy+y^2\right)}}\)
\(\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2y^2+\left(x+y\right)^2}\ge\frac{2x^2}{2x^2+4y^2}+\frac{4y^2}{2x^2+4y^2}=1\)
\(\Rightarrow T\ge1\)
Bài 2:
[Toán 10] Bất đẳng thức | Page 5 | HOCMAI Forum - Cộng đồng học sinh Việt Nam
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 11 2020
Đặt vế trái là P
Ta có: \(P\ge\frac{x^2+1}{1+\frac{y^2+1}{2}+z^2}+\frac{y^2+1}{1+\frac{z^2+1}{2}+x^2}+\frac{z^2+1}{1+\frac{x^2+1}{2}+y^2}\)
Đặt \(\left(x^2+1;y^2+1;z^2+1\right)=\left(a;b;c\right)\Rightarrow a;b;c\ge1\)
\(P\ge\frac{2a}{b+2c}+\frac{2b}{c+2a}+\frac{2c}{a+2b}=2\left(\frac{a^2}{ab+2ac}+\frac{b^2}{bc+2ab}+\frac{c^2}{ca+2bc}\right)\)
\(P\ge\frac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\frac{6\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
19 tháng 8 2020
Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)
KN
13 tháng 3 2021
Theo giả thiết xy + yz + zx = 1 nên ta có: \(VT=\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2}=\frac{1}{xy+yz+zx+x^2}+\frac{1}{xy+yz+zx+y^2}+\frac{1}{xy+yz+zx+z^2}=\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+x\right)\left(y+z\right)}+\frac{1}{\left(z+x\right)\left(z+y\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)Theo bất đẳng thức Cauchy-Schwarz: \(\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^2\le\left(x+y+z\right)\left(\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}\right)=\left(x+y+z\right)\left(\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(y+z\right)\left(y+x\right)}+\frac{z}{\left(z+x\right)\left(z+y\right)}\right)=\frac{2\left(x+y+z\right)\left(xy+yz+zx\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)\(\Rightarrow\frac{2}{3}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)^3\le\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)Ta cần chứng minh: \(\frac{2\left(x+y+z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{4\left(x+y+z\right)}{3\left(x+y\right)\left(y+z\right)\left(z+x\right)}\left(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\right)\)
hay \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}\le\frac{3}{2}\)
Bất đẳng thức cuối đúng theo AM - GM do: \(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{y+z}.\frac{y}{x+y}}+\sqrt{\frac{z}{z+x}.\frac{z}{z+y}}\le\frac{\left(\frac{x}{x+y}+\frac{x}{x+z}\right)+\left(\frac{y}{y+z}+\frac{y}{x+y}\right)+\left(\frac{z}{z+x}+\frac{z}{z+y}\right)}{2}=\frac{3}{2}\)Đẳng thức xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
\(1+y+z^2\le1+\frac{1+y^2}{2}+z^2\)
\(\frac{1+x^2}{1+y+z^2}\ge\frac{2\left(1+x^2\right)}{1+b^2+2\left(1+c^2\right)}\)
Bất đẳng thức cần chứng minh tương đương
\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge1\)
với \(a=1+x^2,b=1+y^2,c=1+z^2\)
\(\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\ge\frac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge1\)
Chứng minh hoàn tất. Đẳng thức xảy ra khi \(a=b=c=1\)
Lâu thiệt lâu mới thấy e ngoi lên á -)))